Which of the following fractions is equivalent to (-13) = (-6)? Select all that apply.

Luna mixes 3/4 cup of orange juice with 3/8 cup of cranberry juice. She gives 5/8 cup of juice to mags. How much is left in luna

Anuta_ua [19.1K]

Answer:

$\frac{1}{2} \: cup \: of \: juice$

Step-by-step explanation:

See the attached picture for explanation.

5 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

Find the value of g. -69 + 36 = 12 HELP ME PLEASE ASAP!!!!

MrMuchimi

Answer:

-6g + 36 = 12

36 = 12 + 6g

24 = 6g

4 = g

g = 4

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

You are using the equation 4(p – 7) = 44 to determine how many pictures can be saved at one time to the photo stream on your cel

OverLord2011 [107]

Answer:

p = 18

Step-by-step explanation:

4(p-7) = 44

The operations in order are to distribute first, 4p - 28 = 44

Isolate the variable, 4p = 72

and divide for the answer, p = 18

P stands for pictures, the number of pictures that can be saved to your photo stream at once is 18.

7 0

2 years ago

Sienna has 5 3/4 pounds of grapes. She eats 1/8 pound and gives 2 1/2 pounds to her brother. How many pounds of grapes does Sien

miskamm [114]

The pounds of grapes left with seinna is 3.125 or $\frac{25}{8}$ pounds

<h3>Solution:</h3>

Given that Sienna has $5\frac{3}{4}$ pounds of grapes

She eats $\frac{1}{8}$ pound and gives $2\frac{1}{2}$ pounds to her brother

To find: pounds of grapes left with seinna

We can find the pounds of grapes left with seinna by subtracting the pounds of grapes she ate and pounds of grapes she gave to brother from the total pounds of grapes she had

Pounds of grapes left with seinna = total pounds of grapes - pounds of grapes she ate - pounds of grapes she gave to brother

$\text {total pounds of grapes }=5 \frac{3}{4}=\frac{4 \times 5+3}{4}=\frac{23}{4}$

$\text {pounds of grapes she ate }=\frac{1}{8}$

$\text {pounds of grapes she gave to brother }=2 \frac{1}{2}=\frac{2 \times 2+1}{2}=\frac{5}{2}$

Substituting the values we get,

$\text {Pounds of grapes left with seinna }=\frac{23}{4}-\frac{1}{8}-\frac{5}{2}$

$\text { Pounds of grapes left with seinna }=5.75-0.125-2.5=3.125$

Thus the pounds of grapes left with seinna is 3.125 or $\frac{25}{8}$ pounds

5 0

3 years ago