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3 years ago

Solve for x. around your answer to the nearest tenth 2.5/10=X/50

Mathematics

2 answers:

Greeley [361]3 years ago

8 0

Answer:

x=12.5

hope this helps:)

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

x = 12.5

Step-by-step explanation:

Cross multiply and solve for x

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Raphael paid $ 492 for a camera during a 25% off sale what was the camera's regular price

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I believe the answer is $1,968

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3 years ago

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Please help me with this math question.

JulijaS [17]

1.SA=10e^2
rectacgular prism=2e*e*e
cube=e*e*e
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3 years ago

Does anyone got the answer to this?

Wittaler [7]

Answer:

Step-by-step explanation:

Givens

C = 2*pi*r

C = pi*d

C = 125.6

r = d/2

Solution

C = pi *d

125.6 = 3.14 * d Divide both sides by 3.14

125.6/3.14 = d

d = 40

You can use 1 of 2 methods to get the radius

The easiest one is use

r = d/2

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2 years ago

Help me please with math

Nady [450]

Answer: There is no question

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3 years ago

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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

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4 years ago