What is 37/15 as a mixed numbers?

An ice cream recipe calls for 6 cups of cream for 7 1/2 gallons. How many cups of cream are needed for 11 1/4 gallons?

denpristay [2]

Answer:

9 cups

Step-by-step explanation:

Since every 6 cups have 7 1/2 gallons, you would divide 7 1/2 gallons by cups which gets you 1.25 as an answer. After you have found out that, then you would divide all 4 of those answers by 11 1/4 gallons, and if your answer is 1.25, well then that is your answer to the question, so it would be 9 cups in this case.

5 0

2 years ago

Read 2 more answers

Solve the inequality for y. y-8x>5

11111nata11111 [884]

Your answer is highlighted in bold at the bottom.
The answer is 5+8x

3 0

3 years ago

Write an equation for the line perpendicular to y=2x 4 that contains (3, -2)?

murzikaleks [220]

In the equation y = 2x + 4, you know that the gradient is 2 (the number before x).

for a perpendicular line you need the negative reciprocal of 2, which is - $\frac{1}{2}$ .

you then need to use the equation
y - $y_{1}$ = m(x - $x_{1}$ )

substitute in your coordinates:

y - -2 = - $\frac{1}{2}$ (x - 3)

simplify:
y + 2 = -0.5x + 1.5

rearrange:

y = - $\frac{1}{2} x$ - $\frac{1}{2}$

3 0

3 years ago

In a sample of 170 students at an Australian university that introduced the use of plagiarism-detection software in a number of

Kisachek [45]

Answer:

$p_v =P(Z>4.146)=0.0000169$

Based on the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .

Step-by-step explanation:

1) Data given and notation

n=170 represent the random sample taken

X=58 represent the student's who belief that such software unfairly targets students

$\hat px=\frac{58}{170}=0.341$ estimated proportion of students who belief that such software unfairly targets students

$\hat p=\frac{112}{170}=0.659$ estimated proportion of students who NOT belief that such software unfairly targets students

$p_o=0.50$ is the value that we want to test

$\alpha=0.05$ represent the significance level (no given)

z would represent the statistic (variable of interest)

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

p= proportion of student's who belief that such software unfairly targets students

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that that a majority of students at the university do not share this belief. :

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p>0.5$

We assume that the proportion follows a normal distribution.

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,higher or less) from a hypothesized value $p_o$ .

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =170*0.5=85>10$

$n(1-p_o)=170*(1-0.5)=85>10$

3) Calculate the statistic

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

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{170}}}=4.146$

4) 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 is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a one right side test the p value would be:

$p_v =P(Z>4.146)=0.0000169$

Based on the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .

8 0

3 years ago

Assume the blood pressures of 10 people were measured before and after sleeping for the night. What is the corresponding critica

NeX [460]

Answer:

For the critical value we need to calculate the degrees of freedom given by:

$df = n-1= 10-1=9$

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

$|t_{crit}| =1.833$

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value with right arm , y = test value with left arm

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

The first step is calculate the difference $d_i=y_i-x_i$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}$

For the critical value we need to calculate the degrees of freedom given by:

$df = n-1= 10-1=9$

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

$|t_{crit}| =1.833$

5 0

3 years ago