All categories

3 years ago

What are the prime factor of 625 a. 25.25 b.1,5,5,5,5 c.5,5,25 d.5,5,5,5

Mathematics

2 answers:

Natalka [10]3 years ago

3 0

d.5,5,5,5. A prime number is a number only divisible by 1 and itself. 1 is not a prime number.

mario62 [17]3 years ago

3 0

   625
       ∧
25   ×   25
   ∧     ∧
5 × 5 × 5 × 5

The answer is D.

You might be interested in

Ethan works in a department store selling clothing. He makes a guaranteed salary of $500 per week, but is paid a commision on to

ElenaW [278]

The total amount of money Ethan made in a week including commission is $812.5

<h3>Total income</h3>

Total sales for the week = $1250
Base salary = $500
Percentage commission = 25%

Amount of commission earned = 25% of $1250

= 25/100 × $1,250

= 0.25 × 1250

= $312.5

Total earnings = Base salary + Amount of commission earned

= $500 + $312.5

= $812.5

Learn more about income:

brainly.com/question/13793671

#SPJ1

3 0

2 years ago

Amanda makes $40,000 and gets an annual raise of $3,000. Marie makes $45,000 and gets an annual raise of $2,500. How many years

pychu [463]

Answer:

Six years

Step-by-step explanation:

because it will divided the twoo digit and mutiple it self

7 0

2 years ago

If shelia earned 36 points for the ducks, how many points did she earn for each frog?

Marina86 [1]

Shealia earned 24 points for the frogs.

6 0

3 years ago

Read 2 more answers

Can someone help me ASAP plsssss

AlladinOne [14]

that is literally gibberish

4 0

3 years ago

A shipment of 12 microwave ovens contains 4 defective units. A restaurant buys three of these units. What is the probability of

hammer [34]

The correct answer is:

20/27 = 0.7407.

Explanation:

To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:

$P(r) = _nC_r(p)^r(1-p)^{n-r}
$

They purchase 3 units total, so this is n.
We want the probability of either 2 or 3 non-defective units; this means r is 2 and r is 3.
In this case we want the probability that the units are non-defective. Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the probability of a non-defective unit 8/12 = 2/3. This is p.

Using these, we have:
$P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3}
\\
\\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0
\\
\\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1)
\\
\\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407$

5 0

4 years ago