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3 years ago

I'm having trouble with a probability problem.

Mathematics

1 answer:

timama [110]3 years ago

6 0

I guess complement should be 1- 0.35 = 0.65 !!

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A survey is being planned to determine the mean amount of time corporation executives watch television. a pilot survey indicated

Natali [406]

Missing question: How many executives should be surveyed?

Solution:
Mean = 13 hours
SD = 3 hours
Confidence level = 95%

Mean viewing time within a quarter of an hour = 0.25 = 1.96*3/sqrt (N)

Where N = Sample population

N = {(1.96*3)/0.25}^2 = 553.19 ≈ 554

Therefore, 554 executives should be surveyed to yield such results.

5 0

3 years ago

Find the area of the shaded region in each of the following figure all unit are in cm

vodomira [7]

1st problem:
A = 630 cm

2nd problem:
A = 104 cm

All I can see on the picture you provided.

7 0

3 years ago

A brick layer is able to set 2.5 breaks in one minute how many bricks can you set an eight hours

frozen [14]

Answer:

<h2> In 8 hours he will set 1200 bricks</h2>

Step-by-step explanation:

In this problem, we are expected to estimate the number of bricks he can set in 8 hours.

let us convert hours to minutes we have

60 min make 1 hour

x min make 8 hours

cross multiply

x= 60*8

x= 480 min

We are assuming the same rate with which he sets 2.5 brick is the same as the rate with which he will set bricks for the 8 hours windows

so if the bricklayer sets

2.5 bricks in 1min

y bricks in 480 min

Cross multiply we have

y=2.5*480

y= 1200 bricks

hence in 8 hours, he will set 1200 bricks

5 0

3 years ago

Javier had $1.30. He wants to split the money between himself and nine friends. Which equation shows how he could divide the mon

Reil [10]

B because it u and 9 friends each want the same amount from splitting 1.30 which gives you .13

3 0

3 years ago

Read 2 more answers

Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.

Georgia [21]

Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

$z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ 
z-score =2.326$

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

5 0

3 years ago