Can someone help answer b plzzz

Answer: A. 7

Step-by-step explanation:

3y-2=19

3y-2+2=19+2 (add 2 to both sides)

3y=21

3y/3=21/3 (divide by 3 on each side)

y=7

Answer:

1000

Step-by-step explanation:

10 x 10 = 100

100 x 10 = 1000

The equation of the ellipse with a major axis 24 units long parallel to the x-axis, a minor axis 22 units long, and a center at (6,1) can be written in the standard form.

The standard form of an ellipse with a major axis parallel to the x-axis is given by: ((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1. where (h, k) represents the center of the ellipse, and a and b represent half the lengths of the major and minor axes, respectively. In this case, the center is (6, 1), so h = 6 and k = 1. The major axis has a length of 24 units, so a = 24 / 2 = 12. The minor axis has a length of 22 units, so b = 22 / 2 = 11. Substituting these values into the standard form equation, we get: ((x - 6)^2 / 12^2) + ((y - 1)^2 / 11^2) = 1.

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Answer:

a < -11

Step-by-step explanation:

Subtract 18 from both sides.

a + 18 - 18 < 7 - 18

= a < -11

The Fraction of the people who spent less than 20 minutes exercising yesterday is; 5/6

Fractions are used to represent the parts of a whole or perhaps the collection of objects. A fraction is seen to have primarily two parts. The number on the top of the line is referred to as the numerator while the number below the line is referred to as denominator.

The total number of people in the survey from the table is;

3 + 22 + 6 + 3 + 1 = 35

Number of people who spent less than 20 minutes exercising yesterday was 25 people.

Thus;

Fraction of those who spent less than 20 minutes exercising yesterday = 25/35 = 5/6

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SOLUTION:

From the diagram, we can see that velocity in component form is;

Answer:

8

Step-by-step explanation:

a. ln(7x) can be expanded as ln(7) + ln(x). b. ln(x + 3) remains as it is, since it cannot be simplified further. c. ln(x^8) can be expanded as 8ln(x). d. 15,000ln(xy^4) can be expanded as ln(x) + 4ln(y) + ln(15,000).

a. To expand the logarithmic expression ln(7x), we can use the property of the natural logarithm that states ln(ab) = ln(a) + ln(b).

Therefore, ln(7x) can be expanded as ln(7) + ln(x).

b. Similarly, the logarithmic expression ln(x + 3) can be expanded using the property ln(ab) = ln(a) + ln(b).

Hence, ln(x + 3) remains as it is since we cannot simplify it further.

c. Expanding the logarithmic expression ln(x^8) can be done using the property ln(a^b) = b * ln(a).

Thus, ln(x^8) becomes 8 * ln(x).

d. Expanding the logarithmic expression 15,000ln(xy^4) can be done by applying the property ln(ab) = ln(a) + ln(b).

Therefore, 15,000ln(xy^4) can be expanded as 15,000[ln(x) + ln(y^4)].

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The probability of at least one of your 10 lottery tickets being a winner is approximately 0.638 or 63.8%.

The probability of winning on a single lottery ticket is 1/8, which means that the probability of not winning is 7/8. If you buy 10 of these lottery tickets, the probability of not winning on any of them is:

(7/8)^10 = 0.362

This means that there is a 36.2% chance that none of your tickets will be a winner. To find the probability that at least one of your tickets will be a winner, we can use the complementary probability:

P(at least one winner) = 1 - P(no winners)

P(at least one winner) = 1 - 0.362

P(at least one winner) = 0.638

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Answer:

7x - 4

Step-by-step explanation:

2x + 4 + 5x - 8

7x + 4 - 8

7x - 4

Answer:

-1/2 anwserchoice B

Step-by-step explanation:

negitive (-) and then rise (1 up) over run (two over)

good morning to you aswell <3

(a) To find the probability that only one machine is in working condition after two days, we need to determine the probability of transitioning from state (1, 1) to state (1, 0) after two days.

From the transition probability matrix, we see that to transition from (1, 1) to (1, 0) in one day, both machines need to remain operational, which has a probability of (1 - p) * (1 - p) = (1 - p)^2.

Therefore, the probability of transitioning from (1, 1) to (1, 0) after two days is ((1 - p) * (1 - p))^2 = (1 - p)^4.

(b) To find the expected number of days until both machines are down, given that currently both machines are operational, we need to consider the transition probabilities from state (2, 0) to state (0, 1).

From the transition probability matrix, we see that to transition from (2, 0) to (0, 1) in one day, both machines need to fail, which has a probability of p * p = p^2.

Therefore, the expected number of days until both machines are down, given that both machines are currently operational, is 1 / (p^2).

(c) To find the steady-state probabilities, we need to solve the equation πP = π, where π is the row vector of steady-state probabilities and P is the transition probability matrix.

Solving this equation will give us the steady-state probabilities for each state (i, j). Since the given matrix is not provided, it is not possible to calculate the exact steady-state probabilities without the specific values of the transition probabilities.

(d) To determine the percentage change in the long-run average benefit per day if p changes from 0.3 to 0.2, we would need to know how the revenue R TL is related to the probability p. Without this information, it is not possible to calculate the percentage change.

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Answer:

1/8 or

1 1/2

Step-by-step explanation:

necesitas contar la xx

Answer:

C) The initial value is 32, and the rate of change is 1.

Step-by-step explanation:

This is the correct answer for edge

Answer:

-2\(\sqrt{3x}\) + 2\(\sqrt{15x}\)

Step-by-step explanation:

The probability that exactly 4 total launches will be performed until an unsuccessful launch occurs is approximately 0.0720 or 7.2%

The probability of exactly 4 total launches being performed until an unsuccessful launch occurs, we can use the geometric probability formula.

In this case, the probability of a successful launch (p) is 0.7 (since the probability of an unsuccessful launch is 0.3). We want to find the probability of 4 successful launches followed by an unsuccessful launch.

The probability of exactly 4 successful launches followed by an unsuccessful launch can be calculated as follows:

P(4 successful launches followed by an unsuccessful launch) = (p⁴) × (1-p)

P(4 successful launches followed by an unsuccessful launch) = (0.7⁴) × (1-0.7)

P(4 successful launches followed by an unsuccessful launch) = 0.7⁴ × 0.3

P(4 successful launches followed by an unsuccessful launch) = 0.2401 × 0.3

P(4 successful launches followed by an unsuccessful launch) ≈ 0.0720

Therefore, the probability that exactly 4 total launches will be performed until an unsuccessful launch occurs is approximately 0.0720 or 7.2%

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The observed allele frequencies for the T102C locus are as follows: T allele frequency = 0.388 and C allele frequency = 0.612.


To calculate the observed allele frequencies, we divide the number of occurrences of each allele by the total number of alleles observed.

In this case, the T allele is observed in 108 TT genotypes and 194 TC genotypes, giving a total of 108 + 194 = 302 T alleles. Similarly, the C allele is observed in 194 TC genotypes and 48 CC genotypes, giving a total of 194 + 48 = 242 C alleles.

To calculate the T allele frequency, we divide the number of T alleles by the total number of alleles: 302 T alleles / (302 T alleles + 242 C alleles) = 0.388 (or approximately 0.39).

Similarly, the C allele frequency is calculated as: 242 C alleles / (302 T alleles + 242 C alleles) = 0.612 (or approximately 0.61).

Therefore, the observed allele frequencies for the T102C locus are T allele frequency = 0.388 and C allele frequency = 0.612.

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Answer:

1

Step-by-step explanation:

Using the Pythagorean theorem, the length of the third side is \(\sqrt{(\sqrt{5})^2-2^2}=1\).

9514 1404 393

Answer:

  A.  Initial value 12, rate of change -2

Step-by-step explanation:

We note that the x-values in the table go up by 1 and the y-values go down by 2. That "down by 2" for "up by 1" is the rate of change: -2/1 = -2.

If we extend the table upward one more row, we have x=0 and y=12. That is, the "initial value" is 12.

The initial value is 12; the rate of change is -2.

Answer:

s=2

Step-by-step explanation:

3+6s=4s-12+19

3+6s=4s+7

6s-4s=7-3

2s=4

Answer:

The rate of change is -0.75

Step-by-step explanation:

For each unit of t the volume of water is decreasing by 0.75.

At t=13.333 the volume of water is 0. at t=0 the volume of water is 10

Hope this helps

Answer:

It has a radius of 4.9meters so its diameter will be multiplied by two making it 9.8meters

Answer: Diameter of each semicircle is 9.8 m, because there is said that these are semicircles, so 4,9 m means radius.

Step-by-step explanation: Semicircle is half of the circle so that means it's divided in half and pass through center point. Radius is the distance from central point to any point of the circumference, and the diameter is equal 2 times Radius.

Answer:

60.89 lbs.

Step-by-step explanation:

(362) x (4/9)

= 1448/9

= 160 8/9 = 60.89 lbs of canned goods

Two examples of functions from Z to Z that are one-to-one but not onto are f: Z → Z, where f(n) is 2n and g: Z → Z, where g(n) is |n|.

Solution: Two examples of functions from Z to Z that are one-to-one but not onto are:

1. f: Z → Z, where f(n) = 2n

This function is one-to-one but not onto because every even integer is in the range of f, but no odd integer is in the range of f.

2. g: Z → Z, where g(n) = |n|.

This function is one-to-one but not onto because every non-negative integer is in the range of g, but no negative integer is in the range of g.

A function f: A → B is said to be one-to-one (or injective) if distinct elements in A have distinct images in B. In other words, if x, y ∈ A and f(x) = f(y), then x is y.

A function f: A → B is said to be onto (or surjective) if every element in B is an image of some element in A. In other words, for every b ∈ B, there exists an a ∈ A such that f(a) = b.

In conclusion, we can have functions that are one-to-one but not onto, which means that there are elements in the codomain that are not images of any element in the domain.

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The countermeasures to avoid potential human errors are   Visual aids & work instructions, education and/or discipline, work standardization and discipline and TPM and skill building etc.

Many major accidents, such as Texas City, Piper Alpha, and Chernobyl, were caused by potential human error. Companies must manage human failure as rigorously as they manage technical and engineering measures to avoid accidents and illness.

There are several countermeasures for amateur or beginner mistakes. Potential human errors can be avoided by some measures including training, skill development, work standardisation, visual aids, and work instructions. Every human being makes mistakes, whether they are professional or not.

Thus, Visual aids and work instructions, education and/or discipline, work standardisation and discipline, TPM and skill building, and other counter measures are used to avoid potential human errors.

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No, there is significant difference in the use of e readers by different age groups.

Given sample 1 ( 29 years old) \(n_{1}\)=628, \(p_{1}\)=7%, sample 2( 30 years old)\(n_{2}\)=2309, \(p_{2}\)=0.11.

We have to first form hypothesis one null hypothesis and other alternate hypothesis.

\(H_{0}:\)π1-π2=0

\(H_{1}:\)π1-π2≠0

α=0.05

Difference between proportions \(p_{1}-p_{2} =-0.04\)

\(p_{d}=0.07-0.11=-0.04\)

The pooled proportion needed to calculate standard error is:

\(p=(X_{1} -X_{2} )/(n_{1} +n_{2} )\)

=(44+254)/(628+2309)

=0.10146

The estimated standard error of difference between means is computed using the formula:

\(S_{p_{1} -p_{2} }=\sqrt{p(1-p)/m_{1}+p(1-p)/n_{2} }\)

=\(\sqrt{0.101*0.899/628+0.101*0.899/2309}\)

=\(\sqrt{0.000143+0.00003}\)

=\(\sqrt{0.000173}\)

=0.01315

Z= Pd-(π1-π2)/\(S_{p_{1} -p_{2} }\)

=-0.04-0/0.013

=-3.0769

This test is a two tailed test so the p value for this test is calculated as (using z table)

p value:2 P(Z<-3.0769)

=2*0.002092

=0.004189

P value< significance level of 5%.

Hence there is enough evidence to show the claim that there is a significant difference in the use of e readers by different age groups.

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Question is incomplete as it also includes:

Significance level of 5%.

two sides are equal means isosceles triangle which means 2 angles are equal, and sum of all angles are = 180, so add 2 angles, subtract them from 180 and u get 3rd angle.

61 + 61 = 122

180-122= 58

x = 58

a) To find the percentage of pregnancies that should last between 26 and 275 days, we can calculate the area under the normal curve between these two values.

Using the standard normal distribution, we need to standardize the values by subtracting the mean and dividing by the standard deviation.

For 26 days:

Z = (26 - 262) / 18 = -12.222

For 275 days:

Z = (275 - 262) / 18 = 0.722

Now, we can find the corresponding probabilities using a standard normal table or a calculator.

The probability of a pregnancy lasting less than 26 days is P(Z < -12.222) which is essentially 0.

The probability of a pregnancy lasting less than 275 days is P(Z < 0.722) = 0.766.

To find the percentage between 26 and 275 days, we subtract the probability of less than 26 days from the probability of less than 275 days:

Percentage = 0.766 - 0 = 0.766 = 76.6%

Therefore, approximately 76.6% of pregnancies should last between 26 and 275 days.

b) To find the number of days for the longest 30% of all pregnancies, we need to find the corresponding Z-score for the upper 30% of the standard normal distribution.

Z(0.30) = 0.524 (approximately)

Now, we can reverse the standardization process to find the corresponding number of days:

X = Z * σ + μ

X = 0.524 * 18 + 262

X ≈ 271.43

Therefore, the longest 30% of all pregnancies should last at least approximately 271.43 days.

c) According to the Central Limit Theorem, the distribution of the sample mean will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

Mean (μ) of the sample mean (y) = Mean of the population = 262 days

Standard deviation (σ) of the sample mean (y) = Standard deviation of the population / √n

σ(y) = 18 / √80 ≈ 2.015

Therefore, the mean of the distribution of the sample mean is 262 days and the standard deviation is approximately 2.015 days.

d) To find the probability that the mean duration of the patients' pregnancies will be less than 200 days, we can standardize the value using the sample mean and standard deviation:

Z = (200 - 262) / (18 / √80) ≈ -7.150

Using a standard normal table or a calculator, we find that P(Z < -7.150) is essentially 0.

Therefore, the probability of the mean duration of the patients' pregnancies being less than 200 days is very close to 0.

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