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3 years ago

4/7b=22 help solve please

1 answer:

7 0

I believe the answer is 77/2. I use this calculator to help me so it might be a little odd of the answer? Need an explanation?

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Does any body know the answer??

Mashcka [7]

Angle V is 40. Angle V and Y are congruent. The angles of a triangle are a sum of 180. Angle V and Angle Y add up to 80. Therefore angle VWZ is 100

8 0

3 years ago

(60)x = 1, what are the possible values of x? Explain your answer.

Gnesinka [82]

Answer:

$\frac{1}{60}$

Step-by-step explanation:

Problem:

The possible values of x in the equation (60)x = 1;

Solution:

Such a problem like this in which the highest power of the unknown is 1 will have just one solution.

60x = 1

To solve this, we take the multiplicative inverse of 60;

the multiplicative inverse of 60 = $\frac{1}{60}$

Use this inverse to multiply both sides of the expression;

$\frac{1}{60}$ $x$ 60x = 1 x $\frac{1}{60}$

x = $\frac{1}{60}$

7 0

3 years ago

Help find the surface area of a cylinder with the height of 9 inches and base radius of 4

dybincka [34]

9514 1404 393

Answer:

104π in² ≈ 326.7 in²

Step-by-step explanation:

Put the given numbers in the surface area formula for a cylinder.

SA = 2πr(r +h)

SA = 2π(4 in)(4 in +9 in) = 104π in² ≈ 326.7 in²

3 0

3 years ago

Read 2 more answers

Help. help. help. help.

Yuri [45]

Answer:

Step-by-step explanation:

8 0

3 years ago

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce

LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that $\mu = 79, \sigma = 22$ .

What percentage of individual adult females have weights between 75 kg and 83 kg?

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$ . So:

X = 83

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?

Now we use the Central Limit THeorem, when $n = 100$ . So $s = \frac{22}{\sqrt{100}} = 2.2.$