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Alenkasestr [34]
3 years ago
8

A T-shirt costs 4 times as much as a pair of socks. a T-shirt costs $16.

Mathematics
1 answer:
alukav5142 [94]3 years ago
3 0
Answer:
D. 4s = 16

Explanation:
We are given that a T-shirt is 4 times the price of the sock.
This means that if we multiply the price of sock by 4, we will get the price of the T-shirt.
In equation form:
price of T-shirt = 4*price of sock

We are given that:
cost of the pair of socks is "s".
cost of T-shirt = 16

Substitute in the above equation, we get:
16 = 4s

Hope this helps :)
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g100num [7]

Answer:

the awnser should be 4.47214 because it's the approximate value of 2√5

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2 years ago
In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ
shepuryov [24]

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

<em><u>Since in this question we are provided with;</u></em>

Sample average weekly food expense, \bar X = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be <u>t-distribution</u> because in the question we don't know anything about population standard deviation (\sigma) .

Normal distribution is used when we know population standard deviation (\sigma).

So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;

                P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Therefore, t-distribution should be used to construct a confidence interval.

3 0
3 years ago
Fruit Company A recently released a new applesauce. By the end of its first year, profits on this product amounted to $37,100. T
Anastasy [175]

Answer: 11 year

P(1) = 37,100

P(4) = 58,400

The linear equation (for x ≥ 1)

P(x) = 37,100 + a(x-1)

For x = 4

58,400 = 37,100 + a(4-1)

58,400 - 37,100 = 3a

21300 = 3a

a = 7100

So, the linear equation:

P(x) = 37100 + 7100*(x-1)

P(x) = 37100 + 7100x - 7,100

P(x) = 7100x + 30000

To find when the profit should reach 108100, we can substitute P(x) by 108100.

108100 = 7100x + 30000

108100 - 30000 = 7100x

78100 = 7100x

x = 78100/7100

x = 11

Answer: 11 year

7 0
1 year ago
Reflect C over the y axis​
4vir4ik [10]

Answer:

C' (- 3, 2 )

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ) , then

C (3, 2 ) → C' (- 3, 2 )

4 0
2 years ago
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

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