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3 years ago

How many cubes with side lengths of 1/2 cm does it take to fill the prism? _______cubes

Mathematics

2 answers:

Dvinal [7]3 years ago

6 0

Answer:

Number of cube fit in prism = 24 cubes

Step-by-step explanation:

Given:

Side of cuboid = 1cm , 2cm , 2 1/2

Find:

Number of cube fit in prism

Computation:

Number of cube fit in prism = Volume of prism / Volume of cube

Number of cube fit in prism = (1)(2)(3/2) / (1/2)³

Number of cube fit in prism = 24 cubes

kati45 [8]3 years ago

4 0

Answer:40 cubes

Step-by-step explanation: i did it on khan academy :D

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3 years ago

Benjamin has 14.2 ounces of water for his plants. He has 10 plants and wants to give each plant the same amount of

frosja888 [35]

Answer:

1.42

Step-by-step explanation:

What you have to do is just divide 14.2 and 10 to get your answer.

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3 years ago

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A veterinarian weighed a sample of 666 puppies. Here are each of their weights (in kilograms): 1, 2, 7, 7, 10, 151,2,7,7,10,151,

rjkz [21]

Answer:

The standard deviation of the weight, σ, is approximately 4.73 kg

Step-by-step explanation:

The weights of the six puppies are;

$x_i$ = 1, 2, 7, 7, 10, and 15

The number of puppies, n = 6

Therefore, we have;

The mean = μ = ∑x/n = (1 + 2 + 7 + 7 + 10 + 15)/6 = 7

The standard deviation, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{n}}$

By substituting, we have;

$\sigma =\sqrt{\dfrac{ \left (1-7 \right )^{2} + \left (2-7 \right )^{2} + \left (7-7 \right )^{2} + \left (7-7 \right )^{2} + \left (10-7 \right )^{2} + \left (15-7 \right )^{2} }{6}}$

Simplifying gives;

$\sigma =\sqrt{\dfrac{ \left (-6 \right )^{2} + \left (-5 \right )^{2} + \left (0 \right )^{2} + \left (0 \right )^{2} + \left (3 \right )^{2} + \left (8 \right )^{2} }{6}} =\sqrt{\dfrac{ 36 + 25 + 0 +0+ 9 + 64 }{6}}$

$\sigma =\sqrt{\dfrac{ 134 }{6}} = \sqrt{\dfrac{67}{3} } \approx 4.72582$

The standard deviation of the weight, σ ≈ 4.73 kg to two decimal places.

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3 years ago

Read 2 more answers

The following hypotheses are given. H0 : π ≤ 0.70 H1 : π > 0.70 A sample of 100 observations revealed that p = 0.75. At the 0

Degger [83]

Answer:

The null hypothesis can't be rejected.

Step-by-step explanation:

Given information:

Null hypothesis:

H₀ : π ≤ 0.70

Alternative hypothesis:

H₁ : π > 0.70

We need to check whether the null hypothesis is rejected or accepted.

If P-value < α, then we reject the null hypothesis H₀.

If P-value ≥ α, then we accept the null hypothesis H₀.

A sample of 100 observations revealed that p = 0.75 at the 0.05 significance level.

Here 0.75>0.05, it means p > α, therefore we can not reject the null hypothesis.

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3 years ago

The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports

german

Answer:

The null and alternative hypothesis are:

$H_0: \mu=5\\\\H_a:\mu< 5$

Test statistic t=-0.256

P-value = 0.4

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that that smartphone carrier's data speeds at airports is less than 5 mbps.

Step-by-step explanation:

The question is incomplete:

Sample mean (M): 4.79

Sample STD (s): 5.8

Sample size (n): 50

This is a hypothesis test for the population mean.

The claim is that that smartphone carrier's data speeds at airports is less than 5 mbps.

Then, the null and alternative hypothesis are:

$H_0: \mu=5\\\\H_a:\mu< 5$

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=4.79.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=5.8.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{5.8}{\sqrt{50}}=0.82$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{4.79-5}{0.82}=\dfrac{-0.21}{0.82}=-0.256$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

This test is a left-tailed test, with 49 degrees of freedom and t=-0.256, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.4) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that that smartphone carrier's data speeds at airports is less than 5 mbps.

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3 years ago