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3 years ago

5

Of the 800 participants in a marathon, 120 are running to raise money for a cause. How many participants out of 100 are running

for a cause

1 answer:

QveST [7]3 years ago

7 0

Im not sure about this question but- 120% of 800 equals- 15%, so 15% could be the answer

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Find the mode of this data set:<br> 31, 30, 32, 31, 30, 31, 32,<br> 33, 30, 31, 31, 30, 32

MariettaO [177]

Answer:

31 is the mode of the data as it appears 5 times

4 0

3 years ago

The first term of a geometric sequence is 300 and the common ratio is 2 What is the 7th term of the sequence

xxMikexx [17]

Step-by-step explanation:

s1 = 300

s2 = s1 × 2 = 300 × 2 = 600

s3 = s2 × 2 = s1 × 2² = 1200

sn = sn-1 × 2 = s1 × 2^(n-1)

s7 = 300 × 2⁶ = 300 × 64 = 19,200

3 0

3 years ago

10 employees only eat

8090 [49]

Answer: the ratio 2:5

Step-by-step explanation:

I think thats the answer,

4 0

3 years ago

What is the x-intercept of y=40x+70

AysviL [449]

At th x-intercept y=o

∴40x + 70 = 0
40x = -70
x = -70/40
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8 0

3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

<u>Part B</u>

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$

<u>Part C</u>

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

6 0

2 years ago