Guys please last one ok

aivan3 [116]

Okay so uh i think it’s this: 15% of 60=0.15* 60=9
9 is what percent of 180 i think..?
9/180=1/20=5/100=5%

7 0

3 years ago

Hershey Park is a popular field trip destination. This year Peirce and Stetson both planned trips there. Peirce rented and fille

serious [3.7K]

A can can hold 18 kids! A bus can hold 58 kids!

3 0

3 years ago

Read 2 more answers

Suppose that a box contains 7 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random without repl

irinina [24]

Answer:

E(x) = 1.43 (Approx)

Step-by-step explanation:

Given:

Total number of camera = 7

Defective camera = 5

Sample selected = 2

Computation:

when x = 0

P(x=0) = 2/7 × 1/6 = 2/42

P(x=1) = [2/7 × 5/6] + [5/7 × 2/6] = 20/42

P(x=2) = 5/7 × 4/6 = 20/42

So,

E(x) = [0×2/42] + [1×20/42] + [2×20/42]

E(x) = 1.43 (Approx)

8 0

3 years ago

Log3(7x) Use the Laws of Logarithms to expand the expression.

Bingel [31]

Log3(7) + log3(x)
Terms that are multiplied inside one log become added when you separate them where each term gets its own log.

7 0

3 years ago

A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than

vovikov84 [41]

Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

$H_0: p \leq 0.05$

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

$H_a: p > 0.05$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that $\mu = 0.05, \sigma = \sqrt{0.05*0.95}$

A random sample of 300 items taken from the production line contained 27 defective items.

This means that $n = 300, X = \frac{27}{300} = 0.09$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}$

$z = 3.18$

Pvalue of the test:

Testing if the mean is greater than a value, which means that the pvalue of the test is 1 subtracted by the pvalue of Z = 3.18, which is the probability of a finding a sample proportion of 0.09 or higher.

Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

5 0

3 years ago