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3 years ago

Help with number 4 please?

Mathematics

2 answers:

LekaFEV [45]3 years ago

5 0

10 because 7 x 3 = 21 and 7 + 3 = 10 so your answer is 10

Elodia [21]3 years ago

5 0

The only factors of 21 are: 1 and 21 or 3 and 7.

Since the two number mentioned are single digits, the numbers being used must be 3 and 7.

To get the sum, just add them together!
3 + 7 = 10

The answer to your problem is 10.

Hope this helps!!

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Someone please help me!!

eduard

On what do you need help

4 0

3 years ago

Ann, Chang, and Kareem sent a total of

Gnoma [55]

It is given in the question that

Ann, Chang, and Kareem sent a total of 89 text messages over their cell phones during the weekend. Ann sent 9 fewer messages than Chang. Kareem sent 3 times as many messages as Ann.

Let number of messages Chang sent be x. And Ann sent 9 fewer than Chang, so Ann sent (x-9) messages. And Kareem sent 3 times as many messages as Ann, so Kareem sent 3(x-9) messages.

And the total messages are 89. That is

$x+x-9+3(x-9)=89$

Combining like terms

$x + x - 9 + 3x-27 =89
\\
5x -36=89
\\
5x = 125
\\
x =25$

So Chang sent 25 messages, Ann sent

$25-9=16 messages$

And Kareem sent

$3*16 = 48 messages$

8 0

3 years ago

What is the degree of the polynomial

stellarik [79]

Binomial is the correct type of polynomial for this problem

3 0

3 years ago

Find the answer The weight of an elephant is 10 to the 3rd power times the weight of a cat .If the the elephant weighs 14,000 po

Hitman42 [59]

Answer:

$14\ pounds$

Below is the procedure that was used to find the answer.

Step-by-step explanation:

Let be "e" the weight in pounds of the elephant and "c" the weight in pounds of the cat.

According to the information provided in the exercise, we know that The weight of an elephant is $10^3$ times the weight of a cat. Based on this we can write the following equation:

$e=10^3c$

If the weight in pounds of the elephant is:

$e=14,000$

We must substitute this value into the equation and then solve for "c" in order to find the weight in pounds of the cat.

Then we get:

$14,000=10^3c\\\\\frac{14,000}{10^3}=c\\\\c=14$

3 0

3 years ago

A die with six faces has the number 1 painted on three of its faces, the number 2 painted on 2 of its faces, and number 3 on one

salantis [7]

Answer:

a) S = {1, 2, 3}

b) P(odd number) = $\frac{2}{3}$

c) No

d) Yes

Step-by-step explanation:

a) The sample space is the set of all possible outcomes. By definition, the elements of a set should not be repeated. Hence, the sample space S = {1, 2, 3}

However, the sample is not equiprobable because each element has different probabilities.

b) P(odd number) = $\frac{number of odd digits}{number of faces}=\frac{4}{6}=\frac{2}{3}$

Note that the odd numbers are 1 (on three faces) and 3 (on one face).

c) The fact the die has been biased does not change the possible outcomes. It only changes the probability of getting any given number.

d) Because the 3-face has been loaded, this probability changes. In fact, it is calculated thus:

Let's assume the probability for 1 or 2 is $x$ . Then that of 3 is $2x$ (because it is twice the others). The sum of probabilities must be 1.

$x+x+x+x+x+2x=1$

$7x=1$

$x=\frac{1}{7}$

P(odd number) = $3\times$ Prob(1) + Prob(3)

= $3\times\frac{1}{7}+\frac{2}{7}$ = $\frac{5}{7}$

7 0

3 years ago