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2 years ago

5

The pet store had 6 puppies selling for $104 each and 12 kittens selling for $24 each. Today on 2 puppies and 8 kittens were lef

t, In dollar terms what was the ration of sales of puppies to kittens

1 answer:

Lunna [17]2 years ago

8 0

You multiply 2*104 that is for puppies and then you do 8 by 24 cause that is for kittens and you get 208:192

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Neko [114]

we are given

$y=-0.0088x^2+0.79x+15$

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y is average gas mileage in miles per gallon

x is speed in miles per hour

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so, we have

$x=60mph$

now, we can plug it and find y

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the best approximate for the average gas mileage at a speed of 60 miles per hour is 30.7 mi/gal...........Answer

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2 years ago

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Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to

RUDIKE [14]

Answer:

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

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

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

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

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

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2 years ago

A customer borrowed $2000 and then a further $1000 both repayable in 12 months. What should he have saved if he had taken out on

Neporo4naja [7]

Answer:

a. $60

Step-by-step explanation:

We will use simple interest formula to solve our given problem.

$A=P(1+rt)$ , where

A= Amount after t years.

P= Principal amount.

r= Interest rate in decimal form.

t= Time in years.

Let us find amount of loans repayable after 12 months for taking two amounts of $2000 and $1000.

As $2000 and $1000 are less than 2500, so the rate of loan will be 10%.

$10\%=\frac{10}{100}=0.10$

12 months = 1 year.

$A=2000(1+0.10\times 1)$

$A=2000(1+0.10)$

$A=2000(1.10)$

$A=2200$

Now let us find amount repayable after 12 months for borrowing $1000.

$A=1000(1+0.10\times 1)$

$A=1000(1+0.10)$

$A=1000(1.10)$

$A=1100$

Adding these amounts we will get total repayable amount after 12 months for borrowing $2000 and $1000 separately.

$\text{Amount repayable for borrowing two separate amounts}=2200+1100=3300$

Now let us find repayable amount after 12 months for taking 1 loan. As $3000 is between $2501 and $7500, so rate of loan will be 8%.

$8\%=\frac{8}{100}=0.08$

$A=3000(1+0.08\times 1)$

$A=3000(1+0.08)$

$A=3000(1.08)$

$A=3240$

Now let us find difference between both repayable loan amounts.

$\text{Difference between both repayable loan amounts}=3300-3240$

$\text{Difference between both repayable loan amounts}=60$

Therefore, the customer should have saved $60, if he had taken out one loan for $3000 and option a is the correct choice.

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lesya [120]

7 MPH because 2 1/3
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jeyben [28]

Answer:

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Step-by-step explanation:

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2 years ago

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