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When making a trail mix, George mixes 51 peanuts with 17 almonds. How many peanuts are in Georges trail mix for every one almond

Ulleksa [173]

Answer:

3

Step-by-step explanation:

51 ÷ 17 = 3

I don't know if I understood the question correctly, but let me know if it was right and it helped! :)

3 0

3 years ago

You cut square corners from a piece of cardboard that has dimensions 32 cm by 40 cm. You then fold the cardboard to create a box

jenyasd209 [6]

Answer:

h = 6 cm

l = (32-2(6)) = 20 cm

w = (40-2(6)) = 28 cm

Step-by-step explanation:

Given in the question a piece of cardboard having dimensions 32 by 40

Suppose we cut square of length x

When you cut square corners from a piece of cardboard, two sides are 32-2x, and other two sides are 40-2x lengths

Formula for the volume of box

v = height * length * width

v = x (32-2x) (40-2x)

v = 4x³ - 144x² + 1280x

Find derivative and equal it to 0

dv/dx = 4(3)x² - (2)144x + 1280

0 = 12x² - 288x + 1280

x1 = 18.110

x2 =5.88

Select x and plug value of x to find dimension

(32-2x) > 0

2x < 32

x < 16

so domain say that x should be less than 16 which means x = 5.88 ≈ 6 cm

h = 6 cm

l = (32-2(6)) = 20 cm

w = (40-2(6)) = 28 cm

6 0

4 years ago

A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.

oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

$H_0: \mu=2.4\\\\H_a:\mu> 2.4$

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58$

The degrees of freedom for this sample size are:

$df=n-1=45-1=44$

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.5801)=0.0066$

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0

3 years ago

Will make Brainly!! Please help its due today

Mazyrski [523]

Answer:

do you have to do all of them

Step-by-step explanation:

did you figure out some of them

7 0

3 years ago

The number 100 has factors 2 and 50, since 2 times 50 = 100. List all the other factor pairs of 100.

jenyasd209 [6]

Answer:

25 and 4

10 and 10

100 and 1

20 and 5

5 0

3 years ago