Pls answer. the question is at the photo

A football player ran the ball for 40 yards, 45 yards, 48 yards, 52 yards, and 105 yards. What is the outlier?

Vinil7 [7]

An outlier is a value that is much higher or much lower than the rest of the values in the data set.

The answer here is 105 because it is nearly double all other values.
65 is incorrect because it's not even a value in the data set.

6 0

4 years ago

What is the simplified form of 8b^3c^ 2a + 4b^3c^2?

stiv31 [10]

$8b^{3} c^2+4b^{3} c^2$
$= (8+4) b^{3} c^2$
$=12 b^{3} c^2$

Be carefully, normally there is no a in your expression. I think you made a mistake.

Hope this helps !

Photon

8 0

3 years ago

Triangle ABC has coordinates A(-2, -3), B(1, 1), and C(2, -1). If the triangle is translated 1 unit right and 3 units up, what a

sleet_krkn [62]

Answer: $A'(-1, 0)$

Step-by-step explanation:

By definition, a Translation is a transformation that moves an object a fixed distance. In Transalation the size, shape and orientation of the object don't change.

The translation of a point $P(x,y)$ "a" units right and "b" units up is:

$P'(x+a,y+b)$

In this case we know that the coordinates of A of the Triangle ABC is A(-2, -3) and the Triangle ABC is translated 1 unit right and 3 units up. Therefore, the the coordinates of A' are:

$A'(-2+1, -3+3)$

$A'(-1, 0)$

7 0

3 years ago

Sarah has a cell phone plan that charges $0.12 per minute plus a monthly fee of $21.00. She budgets $65.50 per month for total c

USPshnik [31]

Answer:

44.5 minutes on her phone to stay within budget

Step-by-step explanation:

65.50-21.00 = 44.50

44.50x0.12 = 5.34

5.34 divided by 0.12 = 44.5

8 0

3 years ago

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in deter

lbvjy [14]

Answer:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.85$ estimated proportion of people who favored candidate A

$p_o=0.8$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.8.:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

8 0

3 years ago