Inverse of y=(x/2)^2

An online health and medicine company claims that about 50% of Internet users would go to an online sight first for

Serggg [28]

Answer:

The claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

Step-by-step explanation:

Since an online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine, and a random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet, to test the claim that less than 50% of Internet users get their health questions answered online, the following calculation must be performed:

1318 = 100

606 = X

606 x 100/1318 = X

60600/1318 = X

45.978 = X

Therefore, the claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

6 0

3 years ago

A car travels at an average speed of 52 miles per hour. How many miles does it travel in 4 hours and 15 minutes?

olchik [2.2K]

221 miles I believe.

6 0

3 years ago

Read 2 more answers

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

3 years ago

Find the inverse of h(x)=x^2 +10. please show work or explain it!!

N76 [4]

Here’s something that might help,when brainly doesn’t help me i use socratic!!!!

6 0

3 years ago

Help pleaseeee???!!!

jolli1 [7]

Answer:

h=8

Step-by-step explanation: $\frac{-7}{8} h= -7\\\\(\frac{8}{-7} )(\frac{-7}{8} h)= -7 (\frac{8}{-7} )\\\\\frac{-56}{-56}h = \frac{56}{7}\\\\ h=\frac{56}{7}\\\\h=8$

the goal is to isolate the variable, get it all by itself. to do that in this problem, we have to move the -7/8 away from the h. we do this by multiplying by the reciprocal (the opposite of the fraction). The reciprocal of -7/8 is 8/-7. So multiply both sides by this and you will get h=8

8 0

3 years ago