Ask question

Ask question

All categories

3 years ago

5

Oliver said, "To the nearest hundred, I’ve attended drama club for 900 days of my life!” Write three numbers that could be the a

ctual number of days Oliver has attended drama club.
A. 743, 970, 933

B. 870, 933, 780,

C. 780, 849, 761,

D. 867, 870, 933

1 answer:

Dmitrij [34]3 years ago

8 0

The answer is D dnsgkstbranat

You might be interested in

Can someone please help? Only answer the question correctly, please!<br><br> Question is down below.

DIA [1.3K]

Answer:

<h2> ${6y}^{2}$ </h2>

Step-by-step explanation:

<em><u>Given</u></em><em><u>, </u></em>

$\frac{1}{ {6y}^{ - 2} }$

<em><u>Since</u></em><em><u>,</u></em>

${x}^{ab} = {( {x}^{a} )}^{b}$

<em><u>Hence</u></em><em><u>,</u></em>

$= \frac{1}{ {( {6y}^{2} )}^{ - 1} }$

<em><u>Since</u></em><em><u>,</u></em>

$\frac{1}{ {x}^{ - 1} } = x$

<em><u>Hence</u></em><em><u>,</u></em>

$= {6y}^{2} (ans)$

7 0

3 years ago

During His hockey career gordie howe played 1767 nhl regular season games scoring 801 goals and making 1049 assists each count a

Fittoniya [83]

Add goals and assists to find the total points:

801 + 1049 = 1850 total points.

Now divide total points by total games to get points per game:

1850 / 1767 = 1.0469

Rounded to nearest hundredth = 1.05 points per game

6 0

3 years ago

Point a(4, 2) is translated according to the rule (x, y) right arrow (x + 1, y – 5) and then reflected across the y-axis.

Alona [7]

A: Quadrant 1

B: The coordinates are 5, -3, quadrant 2

C: -5, -3, in quadrant 3

6 0

3 years ago

PREVIOUS ANSWERS MY NOTES Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minute

rjkz [21]

Answer:

The expected total amount of time in minutes the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n-values of a normal variable:

For the sum of a sample of n values, the mean is of $M = n\mu$ and the standard deviation is of $s = \sigma\sqrt{n}$

Average 2.8 minutes

This means that $\mu = 2.8$

75 calls each day.

This means that $n = 75$

What is the expected total amount of time in minutes the operator will spend on the calls each day?

$M = n\mu = 75(2.8) = 210$

The expected total amount of time in minutes the operator will spend on the calls each day is of 210 minutes.

8 0

3 years ago

The length of time, M hours that a phone battery will work before it needs recharging is normally distributed with a mean of 90

salantis [7]

Answer:

The interquartile range is between 83.25 hours and 96.75 hours

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 90, \sigma = 10$

Find the interquartile range for this distribution.

That is, the middle 50%, from the 25th to the 75th percentile.

25th percentile:

X when Z has a pvalue of 0.25. So X when Z = -0.675

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 90}{10}$

$X - 90 = -0.675*10$

$X = 83.25$

75th percentile:

X when Z has a pvalue of 0.75. So X when Z = 0.675

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 90}{10}$

$X - 90 = 0.675*10$

$X = 96.75$

The interquartile range is between 83.25 hours and 96.75 hours

7 0

3 years ago