How does place value help me subtract decimals?

Round 36 to the nearest hundredth. for the brainiest

mestny [16]

36 to the nearest hundredth is just 36

To round 36 to the nearest hundredth consider the thousandths' value of 36, which is 0 and less than 5. Therefore, the hundredths' value of 36 remains 36

6 0

3 years ago

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60 POINTS + BRAINLIEST

Nitella [24]

Answer:

C

Step-by-step explanation:

By tracing the figure, we see that we go around a semicricle twice and go in a straight path twice. So the perimeter of this figure consists of two semicircles and two line segments, hence the answer is C.

3 0

3 years ago

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write and equation of the line that passes through the given point and is parallel to the given line.1. (4,7), y = 5x-3

Gennadij [26K]

Answer:

y=5x+7

Step-by-step explanation:

parallel lines have the same slope, but different y intercepts

y=5x+7

5 0

3 years ago

Data management: Find n if C ( n+ 2 , 2 ) = 36

Lana71 [14]

Please, Karla, explain what "C" and "n" represent. Are you talking about combinations (for example, n+2 objects taken n at a time? Or is C some kind of function? I don't quite see the relationship of this problem to 'data management.'

3 0

3 years ago

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