How many calories are in 3/4 cup of honey

James purchased 2 T-shirts and a pair of jeans online and paid $80 for his purchase. The next week, the same site was offering a

Maurinko [17]

2t + j = 80
3(0.5)t + 2(0.5)j = 70

j = 80 - 2t
1.5t + j = 70
1.5t + 80 - 2t = 70
-0.5t = 70 - 80
-0.5t = -10
t = -10/-0.5
t = 20

j = 80 - 2t
j = 80 - 2(20)
j = 80 - 40
j = 40

so ur answer is C

6 0

3 years ago

Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 76

lozanna [386]

Using the t-distribution, the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is (8.374, 9.426).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 76 - 1 = 75 df, is t = 1.9921.

The other parameters are given as follows:

$\overline{x} = 8.9, s = 2.3, n = 76$ .

Hence the bounds of the interval are:

$\overline{x} - t\frac{s}{\sqrt{n}} = 8.9 - 1.9921\frac{2.3}{\sqrt{76}} = 8.374$

$\overline{x} + t\frac{s}{\sqrt{n}} = 8.9 + 1.9921\frac{2.3}{\sqrt{76}} = 9.426$

More can be learned about the t-distribution at brainly.com/question/16162795

#SPJ1

4 0

2 years ago

The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The student

Fudgin [204]

Answer:

$z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252$

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas

Step-by-step explanation:

Data given and notation n

n=180 represent the random sample taken

X=157 represent the students who came from Arkansas or a bordering state

$\hat p=\frac{157}{180}=0.872$ estimated proportion of students who came from Arkansas or a bordering state

$p_o=0.9$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 or not than 0.9.:

Null hypothesis: $p\geq 0.9$

Alternative hypothesis: $p < 0.9$

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:

$z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252$

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 $\alpha=0.1$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas

4 0

3 years ago

Please help me with this

ElenaW [278]

Answer:

SAS is the answer ...........

5 0

3 years ago

What is 11% of 15?<br> 1.65<br> 1.56<br> 2.5

Vesna [10]

1.5 +.15 = 1.65. 10% x 15 + 1% x 15

6 0

1 year ago