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3 years ago

10

Toni stacks cube shaped beads that measure 1 centimeter on each edge in a storage box the box can hold six layers of 24 cubes wi

th no gaps or overlaps what is the volume of Tonys storage box

1 answer:

MatroZZZ [7]3 years ago

5 0

The answer is 144 cm. 24*6 equals 144, so that would be the answer

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What is the probability that a five-card poker hand con- tains cards of five different kinds?

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If by kind you mean suits then the probability is zero.

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3 years ago

How do you find a solution for an inequality equation?

kozerog [31]

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Step-by-step explanation:

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3 years ago

Find the missing side lengths. Leave your answer as radicals in simplest form.

Ronch [10]

Answer:

Hence x = y = 2

Step-by-step explanation:

Using the SOH CAH TOA identity

Hypotenuse = 2√2

Opposite = x

theta = 45

Sin theta = opp/hyp

Sin 45 =x/2√2

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x = 2√2 * 1/√2

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y = 2

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3 years ago

Which answer represents a geometric sequence?

exis [7]

Answer:

{−1008,−504,−252,−126,…}

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In this sequence ,we multiply by 0.5 each time to get the next term

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3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ <u><em>proceed with caution</em></u>!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; <u>n < 30</u> for both locations A and B ∴ <u><em>proceed with caution</em></u>!

<u>Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.</u>

<u>Solve for t*:</u>

<u></u>

We need the <u>tail area </u>first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the <u>degree of freedom</u>.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding <u>t-score for df = 5 and tail area = .05</u>.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to <u>interpret</u> the confidence interval of (-20.5707, -11.4293), we can say...

<u><em>We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.</em></u>

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2 years ago