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3 years ago

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A bookstore charges $4 for all children's books. Elijah graphs the function that gives the total cost, y, in dollars to buy x nu

mber of books. Which ordered pair is a point on the graph of the function?
A) (2,8)
B) (4,12)
C) (9,18)
D) (20,5)

2 answers:

Molodets [167]3 years ago

4 0

Answer:

A is the right answer

Step-by-step explanation:

Citrus2011 [14]3 years ago

3 0

A is the right answer

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Devon purchased tickets to an air show for 9 adults and 2

Vladimir79 [104]

Answer:

18$ per adult, 15 per clild

Step-by-step explanation:

Brainly plzzzz

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3 years ago

Read 2 more answers

Please help. Will give brainliest.

evablogger [386]

Alright, so you have the basic formula- good.
You have the A value (400), the interest rate r (7.5% -> .075 in decimal), and the final P value (8500). So, we only need to solve for t.

8500 = (400)(1+.075)^t
/400 /400
21.25 = 1.075^t
logarithms are the inverse of exponents, basically, if you have an example like
y = b^x, then a logarithm inverts it, logy(baseb)=x
Makes sense if you consider a power of ten.
1000 = 10^3
if you put logbase10(1000), you'll get 3.
Anyways, though, to solve the problem make a log with a base of 1.075 in your calculator
log21.25(base 1.075) = t
also, because of rules of change of base (might want to look this up to clarify), you can write this as log(21.25)/log(1.075) = t
Thus, t is 42.26118551.
Rounded to hundredths, t=42.26

6 0

3 years ago

What is the 10th term in the pattern with the formula 9n + 9?

Studentka2010 [4]

A(10) = <span>9(10) + 9= 90+9 =99

---------------------------------------------</span>

6 0

3 years ago

Read 2 more answers

Jane is 5 times older than her sister. In 3 years, Jane’s sister will be 1/4 her age. Find the ages now

Korolek [52]

The present age of Jane is 45 years old and present age of her sister is 9 years old

<em><u>Solution:</u></em>

Let the present age of Jane be "x"

Let the present age of her sister be "y"

<em><u>Jane is 5 times older than her sister</u></em>

present age of Jane = 5(present age of her sister)

x = 5y ---------- eqn 1

<em><u>In 3 years, Jane’s sister will be 1/4 her age</u></em>

Age of sister after 3 years = 3 + y

Age of jane after 3 years = 3 + x

Age of sister after 3 years = 1/4(age of jane after 3 years)

$3 + y = \frac{1}{4}(3 + x)$

Substitute eqn 1 in above equation

$3 + y = \frac{1}{4}(3 + 5y)\\\\12 + 4y = 3 + 5y\\\\5y - 4y = 12 - 3\\\\y = 9$

Substitute y = 9 in eqn 1

x = 5(9)

x = 45

Thus present age of Jane is 45 years old and present age of her sister is 9 years old

3 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago