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2 years ago

One number is 5 more than another. The difference between their squares is 65. What are the numbers?

Mathematics

1 answer:

densk [106]2 years ago

3 0

Answer:

Step-by-step explanation:

Let x be the larger number.

Let y be the smaller number.

x - y = 5

x^2 - y^2 = 65 Factor the left.

(x - y)(x + y) = 65 Substitute the first equation into this equation.

5 *(x + y) = 65 Divide both sides by 5

x + y = 65/5

x + y = 13

Now list the two binomials one after the other.

x + y = 13

x - y = 5 Add

2x = 18 Divide by 2

2x/2 = 18/2

x = 9

x + y = 13 Substitute 9 for x

9 + y = 13 Subtract 9 from both sides.

y = 13 - 9

y = 4

Answer

<em>x = 9</em>
<em>y = 4 </em>

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Manny used 8 tenth-size parts to model 8/10. Ana used fewer parts to model an equivalent fraction. How does the size of a part i

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It should be bigger, think of a pie:
if you cut it into 2 parts, the pieces are big
if you cut it into 4 parts, the pieces are smaller

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3 years ago

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

8 0

3 years ago

8. Hassan made a vegetable salad with 2 3/8 pounds

amm1812

let's convert firstly the mixed fractions to improper fractions and then add up.

$\bf \stackrel{mixed}{2\frac{3}{8}}\implies \cfrac{2\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{19}{8}}~\hfill \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}} \\\\\\ \stackrel{mixed}{2\frac{7}{8}}\implies \cfrac{2\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{23}{8}} \\\\[-0.35em] ~\dotfill$

$\bf \cfrac{19}{8}+\cfrac{5}{4}+\cfrac{23}{8}\implies \stackrel{\textit{using an LCD of 8}}{\implies \cfrac{(1)19+(2)5+(1)23}{8}}\implies \cfrac{19+10+23}{8} \\\\\\ \cfrac{52}{8}\implies \cfrac{13}{2}\implies 6\frac{1}{2}$

5 0

3 years ago