A fish tank is in the shape of a cube. Its volume is 125ft3. What is tge area of one face of the tank?

The ratio of students packing a lunch to those buying a lunch is 2:9. The number of students buying lunch is how many times the

Marta_Voda [28]

4.5 times the number of students packing a lunch

3 0

3 years ago

Read 2 more answers

The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0

4 years ago

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's

Sergeu [11.5K]

Answer:

There is evidence to prove that mean is less than 60 hours.

Step-by-step explanation:

Given that the mean work week for engineers in a start-up company is believed to be about 60 hours.

Sample size taken = 10

$H_0: \bar x=60\\H_a: \bar x$

(left tailed test )

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

$\bar x = 57\\s=6.7823\\se = 0.1448$

Mean difference = -3

Test statistic t =mean differene/std error

= -20.72

p value <0.00001

Since p < 0.05 at 5% level we reject H0

There is evidence to prove that mean is less than 60 hours.

4 0

3 years ago

Find the Area of the shaded region of the trapezoid below. (show your work or explain how you got your answer in words) *

Rudiy27

This is the full answer to the question

5 0

3 years ago

Verify csc^4x-cot^4x=2csc^2x-1

Oksanka [162]

First, factor the left side:

$(csc^2(x))^2-(cot^2(x))^2$

This can be simplified using $x^2-y^2 = (x+y)(x-y)$ where the csc term is x and the cot term is y:

$(csc^2(x)+cot^2(x))(csc^2(x)-cot^2(x))$

One of our Pythagorean Identities states that:

$csc^2(x)-cot^2(x)=1$

So the left side is now:

$csc^2(x)+cot^2(x)$

Now we can use the same Pythagorean Identity to change the right side to:

$2csc^2(x)-(csc^2(x)-cot^2(x))$

All we did was express 1 in terms of csc and cot.

Distribute the negative:

$2csc^2(x)-csc^2(x)+cot^2(x)$

Combine like terms:

$csc^2(x)+cot^2(x)$

Now the left and right sides are the same.

8 0

3 years ago