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2 years ago

8

Your soccor team wants to buy T-shirts.You call two different companies about prices.Each company charges a price per T-shirt an

d a set up fee to create the team logo.The total cost is the same for each company.How many T-shirts is the team buying?

1 answer:

BARSIC [14]2 years ago

5 0

This question is given no background. How big is the soccer teams budget it is to expensive or cheap?

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Q # 6 please solve the equation

nevsk [136]

The answer is 314,000,000 ft^2

pi r^2

3.14* 10,000^2

3.14*100,000,000

314,000,000

6 0

3 years ago

3x - 6y = 27<br><br> 3x + 4y = 17

11111nata11111 [884]

Answer: {x,y} = {27/4,-47/8}

3 0

2 years ago

Read 2 more answers

Factor the four-term polynomial by grouping 3x-3+x^3-4x^2

yaroslaw [1]

(x4−3x3+4x2−8)/(x+1) = x3−4x2<span>+8x−8.</span>

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3 years ago

In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6. Assume the popu

olya-2409 [2.1K]

Answer:

$19.6-2.42\frac{5.8}{\sqrt{42}}=17.43$

$19.6+2.42\frac{5.8}{\sqrt{42}}=21.77$

And the best option for this case would be:

a. (17.5, 21.7)

Step-by-step explanation:

Information given

$\bar X= 19.6$ represent the sample mean

$\mu$ population mean

$\sigma= 5.8$ represent the population deviation

n=42 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=42-1=41$

Since the Confidence is 0.98 or 98%, the significance would be $\alpha=0.02$ and $\alpha/2 =0.1$ , and the critical value would be $t_{\alpha/2}=2.42$

Replacing we got:

$19.6-2.42\frac{5.8}{\sqrt{42}}=17.43$

$19.6+2.42\frac{5.8}{\sqrt{42}}=21.77$

And the best option for this case would be:

a. (17.5, 21.7)

8 0

2 years ago

What can you say about the sum of consecutive odd numbers starting with 1? that is, evaluate 1, 1c3, 1c3c5, 1c3c5c7, and so on,

Mice21 [21]

Let us add consecutive odd numbers and try to find any relationship.

1. 1

2. 1+3 = 4 ( square of 2 i.e $2^{2}$ )

3. 1+3+5 = 9 ( $3^{2}$ )

4. 1+3+5+7 = 16 ( $4^{2}$ )

5. 1+3+5+7+9 = 25 ( $5^{2}$ )

6. 1+3+5+7+9+11 = 36 ( $6^{2}$ )

7. 1+3+5+7+9+11+13 = 49 ( $7^{2}$ )

If we notice, the sum of the consecutive odd integers in each case is equal to the square of the place where it lies. For example, the sum of numbers in seventh place is equal to $7^{2}$ . The sum of the numbers in the fifth line is equal to $5^{2}$ .

4 0

2 years ago