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3 years ago

A survey of 100 people found that 35 people exercise in

Mathematics

1 answer:

Kruka [31]3 years ago

8 0

Answer:

Jim's claim is correct

Step-by-step explanation:

Total survey=100 people

Morning exercise=35 people

Afternoon exercise=45 people

Night exercise=20 people

Morning exercise : Total survey

=35 : 100

Afternoon exercise : Total survey

=45 : 100

Night exercise : Total survey

=20 : 100

Jim's claim is correct because in finding the ratio, one has to relate it with the total people involved in the survey.

Tara's claim however seems to relate one variable with two other variables

Example:

35 : 55

Morning exercise : (morning exercise + night exercise)

35 : (35+20)

35 : 55

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Find an equivalent fractions for each given fraction 4/10

Harman [31]

The answer would be 2/5

4. divided by 2 = 2
—-
10. divided by 2 = 5

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3 years ago

A cone has a volume of 314 cubic millimeters and a height of 12 millimeters. What is its

Tom [10]

Answer:

The answer is 14

Step-by-step explanation:

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3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution 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.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

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

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

3 years ago

Point P has coordinates P(-4, 3).

kipiarov [429]

YXZ equals P on the side of VWVP equals W

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2 years ago

Karen bakes cookies. She has observed that the number of cookies that turn out well in each batch varies with the size of the ba

mestny [16]

Substitute the values of the size of the 5 batches to the model.

For example

@x = 20

f(20) = 2*20 – 23 = 17

f(25) = 2*25 – 23 = 27

f(26) =29

f(27) = 31

f(34) = 45

Therefore the answer is the first choice (17, 27, 29, 31, 42)

6 0

4 years ago