All categories

3 years ago

What are the prime factors of 625?

Mathematics

2 answers:

Alika [10]3 years ago

5 0

The answer is B! Because 5 x 5 x 5 x 5 = 625

Flura [38]3 years ago

4 0

All of the choices are Factors of 625

But, the prime factors of 625 are 5,5,5,5 because prime numbers are natural numbers greater than 1 that has no positive divisors other than 1 and itself.

Therefore, the answer is Letter B

You might be interested in

Orly uses 2 cups of raisins for every 8 cups of trail mix she makes. How many cups of trail mix will she make is she uses 12 cup

allsm [11]

You have a 2:8 ratio, or 1:4. Since you have 12 cups of raisins, you're going to get the answer 48.

8 0

3 years ago

Read 2 more answers

The base of a rectangle is 13 centimeters longer than the height.If the perimiter is 58 centimeters find the base and the height

Yuri [45]

Answer:

Base: 8cm. Height: 21cm

Step-by-step explanation:

We know that the base of a rectangle is 13 cm. longer than the height and the perimeter is 58 cm. Here is how we would set up the problem:

2x+2(x+13)= 2x + 2x + 26 = 4x + 26 = 58 (now we use the distributive property:

4x + 26 = 58

-26 -26

4x = 32

So, the base is 8 cm.

Now it's time to find the height.

So, we know the base (in total) is 16 cm. Now we do 58-16=42. Now we divide 42 divided by 2 = 21 cm.

8 0

3 years ago

A solid metal rectangular pyramid has a base that measures 16 cm by 25 cm The Solid is melted and recast into a solid cavare pyr

Ymorist [56]

12 yesyrsyrsjsnskwnbewnisenenekek

5 0

3 years ago

Rectangle A measures 10 inches by 8 inches. Rectangle B is an enlarged copy of Rectangle A. Select all of the measurement pairs

Fudgin [204]

Answer:

See Explanation

Step-by-step explanation:

Given

Rectangle A:

$Length: 10\ in$

$Width: 8\ in$

Required

Determine the possible dimensions of Rectangle B

The question has missing options; however, the question can still be solved.

Rectangle B being an enlarged copy of A implies that the dimension of A A is enlarged in equal proportion to form B

Take for instance, the measurements of Rectangle B are

$Length: 15 in$

$Width: 12 in$

Divide the corresponding lengths of B by A to get the enlargement ratio

$Ratio = \frac{B}{A}$

For length:

$Ratio = \frac{15}{10}$

$Ratio = 1.5$

For width

$Ratio = \frac{12}{8}$

$Ratio = 1.5$

Notice that both ratios are the same.

For this measurement of B, we can conclude that B is an enlargement of A

Assume another measurements for B

$Length: 20\ in$

$Width: 10\ in$

Calculate Ratios

$Ratio = \frac{B}{A}$

For length:

$Ratio = \frac{20}{10}$

$Ratio = 2$

For width

$Ratio = \frac{10}{8}$

$Ratio = 1.25$

Notice that, both ratios are not equal.

For this measurement of B, we can conclude that B is not an enlargement of A

Conclusively, all you have to do is: determine the ratios of the dimensions of B to A, if the result are equal then B is an enlarged copy of A; if otherwise, then B is not an enlarged copy

5 0

3 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago