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2 years ago

5

According to a survey of workers 6/50 of them walk to break 3/50 bike 10/50 carpool and 31/50 Drive alone what percent of worker

s walk or bike to work

1 answer:

otez555 [7]2 years ago

7 0

Answer:

Step-by-step explanation:

6+3=9/50

9/50 x 100=18 percent

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Find the circumference of the object. Use $3.14 or 22/7

kodGreya [7K]

Answer:

18.84

Step-by-step explanation:

Circumference = 2πd

where d = diameter

the shape given has a diameter of 6

to find the circumference we simply plug in the value of the diameter as well as π into the formula

formula : C = πd

π = 3.14 and d = 6

C = 3.14(6) = 18.84cm

5 0

2 years ago

According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m

tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

$P(A) = 0.8x + 0.25*0.2x = 0.85x$

Probability of developing diabetes while being very active:

0.25x of 0.2. So

$P(A \cap B) = 0.25x*0.2 = 0.05x$

What is the probability that a middle-aged man with diabetes is very active?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588$

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0

3 years ago

The steps to construct a segment bisector to a given line segment using a compass and straightedge are given below. Arrange them

swat32

4, 1, 5, 2, 3

The appropriate order is ...
(4) Open a compass so its width is more than half the length of the given segment.

(1) Without changing its width, use the compass to draw an arc above and below the given line segment from one of the segment"s endpoints.

(5) Again, without changing its width, use the compass to draw another pair of arcs from the other endpoint. One arc will be above the segment while the second arc will be below.

(2) Draw the point of intersection between the pair of arcs above the line segment and between the pair of arcs below the line segment.

(3) Use a straightedge to connect the intersection points between each pair of arcs.

6 0

3 years ago

Simplify the following<br> (-3a^2b^6)^2

JulsSmile [24]

Answer:

$9a^{4} b^{12}$

(or)

9a^4b^12

6 0

2 years ago

A: {71,73,79,83,87} B:{57,59,61,67}

Jobisdone [24]

Answer:

$\frac{3}{5}$ .

Step-by-step explanation:

We have been given two sets as A: {71,73,79,83,87} B:{57,59,61,67}. We are asked to find the probability that both numbers are prime, if one number is selected at random from set A, and one number is selected at random from set B.

We can see that in set A, there is only one non-prime number that is 87 as it is divisible by 3.

So there are 4 prime number in set A and total numbers are 5.

$P(\text{Prime number from A})=\frac{4}{5}$

We can see that in set B, there is only one non-prime number that is 57 as it is divisible by 3.

So there are 3 prime number in set B and total numbers are 4.

$P(\text{Prime number from B})=\frac{3}{4}$

Now, we will multiply both probabilities to find the probability that both numbers are prime. We are multiplying probabilities because both events are independent.

$P(\text{Prime number from A and B})=\frac{4}{5}\times \frac{3}{4}$

$P(\text{Prime number from A and B})=\frac{1}{5}\times \frac{3}{1}$

$P(\text{Prime number from A and B})=\frac{3}{5}$

Therefore, the probability that both numbers are prime would be $\frac{3}{5}$ .

4 0

3 years ago