Task 4 Task 다. Given: m

1. Mr. McClellan compared the weights (in pounds) of pairs of elk antlers dropped at Mount St Helens NVM and Rocky Mountain NP.

Dahasolnce [82]

(a) I have the line plot attached
(b) Purple mean: 30
Red mean: 36
Purple median: 30
Red median: 36

I'm sorry, that is as much as I can help with, I really am sorry.

4 0

4 years ago

Please answer the question in the photo, thanks!

belka [17]

Answer:

$\therefore 2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right ) = -\dfrac{7}{11}$

Step-by-step explanation:

The question relates to dividing a fraction by proper and another fraction

The given expression is presented as follows;

$2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right )$

We rearrange the fractions as improper fractions for easier division as follows;

$2\dfrac{1}{3} = \dfrac{7}{3}$

$-3\dfrac{2}{3} = -\dfrac{11}{3}$

Therefore, we have;

$2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right ) = \dfrac{7}{3} \div \left (-\dfrac{11}{3} \right ) = \dfrac{7}{3} \times \left (\dfrac{1}{-\dfrac{11}{3} } \right ) = \dfrac{7}{3} \div \left (-\dfrac{3}{11} \right ) =-\dfrac{7}{11}$

$\therefore 2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right ) = -\dfrac{7}{11}$

4 0

3 years ago

An English professor assigns letter grades on a test according to the following scheme. A: Top 14% of scores B: Scores below the

NeX [460]

Answer:

Grades between 62 and 64 result in a D grade.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 71.9, \sigma = 7.8$

Find the numerical limits for a D grade.

D: Scores below the top 84% and above the bottom 10%

So below the 100-84 = 16th percentile and above the 10th percentile.

16th percentile:

This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.

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

$-0.995 = \frac{X - 71.9}{7.8}$

$X - 71.9 = -0.995*7.8$

$X = 64$

10th percentile:

This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

$-1.28 = \frac{X - 71.9}{7.8}$

$X - 71.9 = -1.28*7.8$

$X = 62$

Grades between 62 and 64 result in a D grade.

6 0

3 years ago

Please help me ASAP it’s times THANK U

goblinko [34]

Answer:

Only choices C and D are solutions

Step-by-step explanation:

6x + 3y = -15

y = -2x - 5

6x + 3y = -15

6x + 3(-2x - 5) = -15

6x - 6x - 15 = -15

0 = 0

Since 0 = 0 is a true statement, both equations of this system are the same equation and represent a single line on the coordinate plane.

We need to check each choice in just one equation.

Let's use the second equation.

y = -2x - 5

A.

(2, 7)

7 = -2(2) - 5

7 = -4 - 5

7 = -9 False

Not a solution

B.

(5, 0)

0 = -2(5) - 5

0 = -10 - 5

0 = -15 False

Not a solution

C.

(-3, 1)

1 = -2(-3) - 5

1 = 6 - 5

1 = 1 True

Solution

D.

-13 = -2(4) - 5

-13 = -8 - 5

-13 = -13 True

Solution

Answer: Only choices C and D are solutions

7 0

3 years ago

Two chemists working for a chicken fast-food company, have been producing a very popular sauce. Let’s call then Jesse and Mr. Wh

a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

Let $\overline{x}$ and $\overline{y}$ be sample means of white and Jesse denotes are two random variables.

Given that both samples are having normally distributed.

Assume $\overline{x}$ having with mean $\mu_{1}$ and $\overline{y}$ having mean $\mu_{2}$

Also we have given the variance is constant

A)

We can test hypothesis as

$H0: \mu_{1} = \mu_{2}H1: \mu_{1} > \mu_{2}$

For this problem

Test statistic is

$T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}$

Where

$s=\sqrt{\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}}$

We have given all information for samples

By calculations we get

s=2.41

T=2.52

Here test statistic is having t-distribution with df=(10+7-2)=15

So p-value is P(t15>2.52)=0.012

Here significance level is 0.05

Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.

We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.

3 0

3 years ago

Task 4 Task 다. Given: m​

Task 4 Task 다. Given: m