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3 years ago

Draw 10 marbles use 5-group.

Mathematics

2 answers:

Tatiana [17]3 years ago

6 0

2 cuz 10 dived by 5

LenaWriter [7]3 years ago

4 0

Answer:

Step-by-step explanation:

there would be 2 in every group, 10 divided by 5

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Y= what ? Need help ASAP

lana66690 [7]

The equation is asking for you to write out an equation y= x + b. The equation should be y= 2x - 1

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3 years ago

Read 2 more answers

Determine the greatest common factor of 15xy^7z and 30x^2y^7z

yawa3891 [41]

Answer:

$15x {y}^{7} z$

Step-by-step explanation:

<h3>Factorize the Expressions</h3>

$15x {y}^{2} z = 3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z$

$30 {x}^{2} {y}^{7} z = 2 \times 3 \times 5 \times x \times x \times y \times y \times y \times y \times y \times y \times y \times z$

<h3>Factor out the Common Factors</h3>

$gcf = 3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z$

<h3>Multiply the Common Factors</h3>

$3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z = 15x {y}^{7} z$

The Factored expressions are

$15x {y}^{7} z$

And

$15x {y}^{7} (2x)$

5 0

3 years ago

Using the diagram above,what is the m

makkiz [27]

I’m pretty sure the angle in 53 degrees

4 0

3 years ago

A pen costs p cents. Write down a formula for the cost, C cents, for n pens.

Aleks [24]

Answer:

n x p = c

Step-by-step explanation:

n is the amount of pens so we time it by the cost of 1 pen which is p and the final answer is c as the cost of all the pen.

6 0

3 years ago

Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin

Artemon [7]

"<span>Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find $\bar{d}$ , $s_{d}$ , the t-test statistic, and the critical values to test the claim that $\mu_{d} = 0$ "

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find </span><span> $\bar{d}$
You are asked to find the mean difference between the two variables, which is given by the formula:
$\bar{d} = \frac{\sum (x - y)}{n}$

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example:
</span><span> $\bar{d} = \frac{6}{8} = 0.75$ </span>

B) Find <span> $s_{d}$
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span> $s_{d} = \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }$

These are the steps to follow:
1) Subtract the mean difference from each pair's difference
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
$s_{d} = \sqrt{ \frac{101.5}{8-1} }$
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
$t = \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}$

where SE = standard error is given by the formula:
$SE = \frac{ s_{d} }{ \sqrt{n} }$

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract $\mu_{d}$
6) divide by the standard error

In our example:
SE = 3.81 / √8
= 1.346

The problem gives us <span> $\mu_{d} = 0$ , therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
= 0.56

D) Find $t_{\alpha / 2}$
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find <span> $t_{\alpha / 2} = 1.895$ </span>

Since our t-value is less than <span> $t_{\alpha / 2}$ </span> we can reject our null hypothesis!!

7 0

3 years ago