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Radda [10]
3 years ago
14

Which two of the prime numbers below are factors of 247. a.11 b.13 c.17 d.19

Mathematics
1 answer:
guapka [62]3 years ago
8 0

a) 247 ÷ 11 = 22 r.5    <em>since there is a remainder it is not a factor</em>

b) 247 ÷ 13 = 19      there is no remainder so this is a factor!  The other factor is 19.

Answer: B, D

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Help me plz. I dont get this
Ulleksa [173]
Thats easy. figure it out

7 0
3 years ago
3. Peri earned $27 for walking her neighbor's dog 3 times. If pe the same rate and earned $36, how many times did she wall neigh
Andrei [34K]

Answer:

She walked neighbor's dog 4 times.

Step-by-step explanation:

It is given that:

Peri earned $27 for walking the dogs 3 times.

Unit rate = \frac{27}{3}

Unit rate = $9 per dog

Now,

Peri worked at the same rate and earned $36.

Unit rate = \frac{Total\ earning}{Dogs\ walked}

9 = \frac{36}{d}

d= 36/9

d = 4

Therefore,

She walked neighbor's dog 4 times.

7 0
3 years ago
Three consecutive integers add up to 51. What are these integers​
kaheart [24]

Let the three consecutive integers be x, x+1 and x+2.

According to the question, x+x+1+x+2=51

=>3x+3=51

=>3x+3-3=51-3

=>3x=48

=>x=16

Hence, first integer = 16,

second integer = 16 + 1 = 17 and

third integer = 16 + 2 = 18.

3 0
3 years ago
- Ashanti earns $50 an hour. How much does she earn in 6 hours?
Vesnalui [34]

$50×6=$300

(This is direct proportion question)

(Dependent variable is money earned)

(Independent variable is hours worked)

3 0
3 years ago
Read 2 more answers
Astronomers treat the number of stars X in a given volume of space as Poisson random variable. The density of stars in the Milky
cupoosta [38]

Answer:

Explained below.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of stars in a given volume of space.

X\sim \text{Poisson}\ (\lambda=1)

The probability mass function of <em>X</em> is:

p_{X}(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}

(7)

Compute the probability of exactly two stars in 16 cubic light-years as follows:

P(X=2)=\frac{e^{-1}\times 1^{2}}{2!}=\frac{e^{-1}}{2}=\frac{0.36788}{2}=0.18394\approx 0.184

(8)

Compute the probability of three or more stars in 16 cubic light-years as follows:

P(X\geq 3)=1-P(X

(9)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in 4 cubic light years there will be, 4 × (1/16) = (1/4) stars.

(10)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in <em>t</em> cubic light years there will be, [<em>t</em> × (1/16)] stars.

7 0
3 years ago
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