Help me please 30 points

A model is 4.2 feet tall, the plywood used to make the model is 0.6 inch thick. How many sheets of plywood to make the model?

nadya68 [22]

Answer=84 sheets

4.2feet
1 foot=12 inches

4.2*12=<span>50.4inches

50.4/0.6=</span><span>84 sheets</span>

4 0

3 years ago

A college statistics class at a nearby university has 1,000 students enrolled in it. On a recent exam, the mean score was 75% wi

IceJOKER [234]

Answer:

38.11%

Step-by-step explanation:

Given that:

Mean (μ) = 75, standard deviation (σ) = 5

Z score is a measure in statistics to determine the variation of a raw score from the mean. It is given by the equation:

$z=\frac{x-\mu}{\sigma}$

To calculate the percentage of students scored between a 73 and 78 (C grade), we need to find the z score for 73 and then for 78.

For x = 73, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{73-75}{5} =-0.4$

For x = 78, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{78-75}{5} =0.6$

From the probability distribution table:

P(73 < x < 78) = P(-0.4 < z < 0.6) = P(z < 0.6) - P(z < -0.4) = 0.7257 - 0.3446 = 0.3811 = 38.11%

4 0

3 years ago

Help! Match each linear equation with its graph

Bond [772]

Answer:

y = -1/4 x+4 is blue

y = 1/2 x+4 is green

y = 3x is red

y = -1/2x is purple

y = 3 is black

6 0

3 years ago

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

valkas [14]

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :

Null hypothesis: $\mu \geq 2$

Alternative hypothesis: $\mu < 2$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0

3 years ago

In answering a question on a multiple-choice test, a student either knows the correct answer or guesses it. Let 0.7 be the proba

Dahasolnce [82]

Answer:

The probability is 0.9211

Step-by-step explanation:

Let's call K the event that the student know the answer, G the event that the student guess the answer and C the event that the answer is correct.

So, the probability P(K/C) that a student knows the answer to a question, given that she answered it correctly is:

P(K/C)=P(K∩C)/P(C)

Where P(C) = P(K∩C) + P(G∩C)

Then, the probability P(K∩C) that the student know the answer and it is correct is:

P(K∩C) = 0.7

On the other hand, the probability P(G∩C) that the student guess the answer and it is correct is:

P(G∩C) = 0.3*0.2 = 0.06

Because, 0.3 is the probability that the student guess the answer and 0.2 is the probability that the answer is correct given that the student guess the answer.

Therefore, The probability P(C) that the answer is correct is:

P(C) = 0.7 + 0.06 = 0.76

Finally, P(K/C) is:

P(K/C) = 0.7/0.76 = 0.9211

7 0

3 years ago