PLEASE HELP! I will give brainliest!!

HELLLLPPPPP PLZZZZ!!!!

motikmotik

Answer:

width = 5 ft

Step-by-step explanation:

1.9 times of width = length

1.9*w = 9.5 ft

1.9w = 9.5

Divide both sides by 1.9

w = 9.5/1.9

w = 5 ft

6 0

3 years ago

Read 2 more answers

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

4 years ago

A bag contains 10 marbles,7 of which are black and 3 are red.three marbles are drawn at one after the other.find the probability

S_A_V [24]

<h3>Answer: 17/24</h3>

==========================================================

Explanation:

We have these four cases or possible outcomes

Case 1) We select 0 black marbles and 3 red marbles.
Case 2) We select 1 black marble and 2 red marbles.
Case 3) We select 2 black marbles and 1 red marble.
Case 4) We select 3 black marbles and 0 red marbles.

Let's calculate the probability for case 4.

There are 7 black marbles out of 10 total. The probability of picking black is 7/10. If no replacement is made, then 6/9 is the probability of picking black again (subtract 1 from the numerator and denominator separately). Finally, 5/8 is the probability of getting black a third time.

The probability of getting 3 black marbles in a row is

(7/10)*(6/9)*(5/8) = (7*6*5)/(10*9*8) = 210/720 = 7/24.

That fraction 7/24 means that if you had 24 chances, then you expect about 7 of them will lead to getting three black marbles in a row (aka case 4). Therefore, 24-7 = 17 occurrences are expected where we get cases 1 through 3 occur in some fashion (pick one case only).

Notice how cases 1 through 3 encapsulate the phrasing "at most 2 black marbles" which is another way of saying "2 black marbles is the highest we can go".

So that's why the answer is 17/24.

8 0

3 years ago

A study was performed with a random sample of 3000 people from one church. What population would be appropriate for generalizing

Reil [10]

The population would be set of all Christians of that particular church from where the random samples is collected.

According to the questions,

A study was performed with a random sample 3000 people from one church. In order to find appropriate population for generalizing conclusions

The population would be set of all Christians of that particular church from where the random samples is collected.

This is because random samples is collected only from the particular church, so its generalization can only be the possible to large population of that church.

we can generalize to whole of that church because it is given that there was no bias is introduced.

Hence, the population would be set of all Christians of that particular church from where the random samples is collected.

Learn more about random samples here

brainly.com/question/17040012

#SPJ4

7 0

2 years ago

Please help explanation if possible

GREYUIT [131]

Answer:

(x,y) —> (–1 ,1)

I hope I helped you ^_^

3 0

3 years ago