A survey asked adults how often they exercised and where they most often exercised. The results are shown in the table.

Mathematics

2 answers:

Lera25 [3.4K]2 years ago

7 0

Answer:

First blank: 46%; second blank: 54%

Step-by-step explanation:

There are a total of 92+110 = 202 people who exercise 3 times per week.

Out of these, 92 exercise outdoors; this gives us

92/202 = 0.455 ≈ 0.46 = 46%

110 of these exercise in a gym; this gives us

110/202 = 0.5446 ≈ 0.54 = 54%

Ne4ueva [31]2 years ago

3 0

Your answer is of those who excercise 3 or more times per week, about 46% usually excersise outdoors and about 54% usually excercise in a gym.

hope I helped have a great day!!

You might be interested in

Hong Tang's gross weekly salary is $615. His weekly federal withholding is $55.

Ulleksa [173]

too much math for my mental health

7 0

2 years ago

A survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This

podryga [215]

Answer:

test statistic (Z) is 2.5767 and p-value of the test is .009975

Step-by-step explanation:

$H_{0}$ : percentage of students who smoke did not change

$H_{a}$ : percentage of students who smoke has changed

z-statistic for the sample proportion can be calculated as follows:

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of smoking students ( $\frac{50}{200}$ =0.25)

p is the proportion of smoking students in the survey conducted five years ago (18% or 0.18)
N is the sample size (200)

Then, z= $\frac{0.25-0.18}{\sqrt{\frac{0.18*0.82}{200} } }$ ≈ 2.5767

What is being surveyed is if the percentage of students who smoke has changed over the last five years, therefore we need to seek two tailed p-value, which is .009975.

This p value is significant at 99% confidence level. Since .009975 <α/2=0.005, there is significant evidence that the percentage of students who smoke has changed over the last five years

5 0

2 years ago

A bowl of flower seeds contains 5 petunia seeds and 15 begonia seeds. Riley calculated the probability that a random selected se

Vlad1618 [11]

Answer:

Riley divided the number of petunia seeds by the number of begonia seeds, getting $\frac {5}{15} = \frac {1}{3}$ . What you actually need to do to calculate the probability of selecting a petunia seed is divide the number of petunia seeds by the <em>total</em> number of seeds. 5 petunia seeds and 15 begonia seeds makes 20 total seeds, so you divide 5 petunia seeds by 20 total seeds to get $\frac {5}{20} = \frac {1}{4}$ .