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3 years ago

15

Peaches are being sold for 2$ per pound. If x represents the number of pounds of peaches bought and y represents the total cost

of the peaches, which best describes the values of x and y

1 answer:

ladessa [460]3 years ago

6 0

3x-x+2=4, That should be your answer if that is what your looking for

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two pipes together can fill a tank in 12 hours alone the smaller pipe can take 7 hours longer than the larger pipe takes to fill

Rashid [163]

Well you should just have to add 7 to 12.

3 0

3 years ago

What is the area of the surface of a cookie was a diameter of 32 feet?

aliya0001 [1]

That would be $\pi(32/2)^2\approx804.25\mathrm{ft^2}$ .

Hope this helps.

4 0

3 years ago

MAXImum [283]

Answer:

$32.39

Step-by-step explanation:

She has a 15-gallon tank and it has 1/5 a tank of gas left (so about 3/15)

Because she needs to refill it and already has 1/5, she'll only need to pay for 4/5 a tank of gas (12/15).

So the equation should be 12 times $2.699.

12 × 2.699 = 32.388

Then, round to the nearest cent.

32.388 = $32.39

Hope this helped !

5 0

2 years ago

Position to term rule of 2, 6, 10, 14, 18<br><br> multiply by _____ and subtract by _____

andreyandreev [35.5K]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the sequence

2, 6, 10, 14, 18

An arithmetic sequence has a constant difference and is defined as

$\:a_n=a_1+\left(n-1\right)d$

compute the differences of all the adjacent terms

$6-2=4,\:\quad \:10-6=4,\:\quad \:14-10=4,\:\quad \:18-14=4$

The difference between all the adjacent terms is the same.

Thus,

$d=4$

and

$a_1=2$

Therefore, the nth term is computed by:

$a_n=4\left(n-1\right)+2$

$a_n=4n-2$

Thus, position to term rule of 2, 6, 10, 14, 18 multiply by __4___ and subtract by __2__.

4 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago