A pet store has a total of 45 cats and dogs. There are 7 more cats than dogs. Find the number of cats and dogs in the store.

Mathematics

1 answer:

shutvik [7]4 years ago

7 0

Their are 26 cats and 19 dogs in the store . 26+19=45 .

You might be interested in

What is 4/7 + 2/7 in simplest form

Arada [10]

$\frac{2}{7} +\frac{4}{7}\\
\frac{2+4}{7}\\
\frac{6}{7}$
Your answer is $\frac{6}{7}$

3 0

3 years ago

Write a<br> rule that could be used to simplify the square of any square root of a number.

Vlada [557]

Product rule- the first rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions

HOPES THIS HELP :)

5 0

3 years ago

The volume of a cube is 125 cubic inches. What is the length of one side of the cube? A. 5 inches B. 25 inches C. 125 inches D.

amid [387]

My answer would be B! Hope this helps

4 0

3 years ago

You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the a

svetoff [14.1K]

Answer:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

Step-by-step explanation:

We have the following info:

$ME = 50$ margin of error desired

$\sigma = 1000$ the standard deviation for this case

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =50 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=0.01$ . And for this case would be $z_{\alpha/2}=2.58$ , replacing into formula (b) we got: