All categories

3 years ago

I need Wolfkid1563 to see this. this is the problem i am having trouble with.

Mathematics

2 answers:

DerKrebs [107]3 years ago

8 0

The answer is 5 Ibs I took the test

MrRissso [65]3 years ago

5 0

Answer:

i believe that the answer is 5 lbs

Step-by-step explanation:

i hope this is right and i am sorry if it is wrong and i doubt that Wolfkid1563 will see this unless u message him and tell him unless u dont have enough points

You might be interested in

PLEASE HELP DIRECTIONS INCLUDED IN PHOTOS.

son4ous [18]

I just took the test yesterday and passed
1. First one is False second one is true
2. the points are -2,-3 1,6 2,9
3. -3
4. 1

5 0

3 years ago

Read 2 more answers

Pls show work will mark brainliest

maw [93]

The answer would be B.52 because, when you divide 104 by 78 you get about 1.333333. Which is the number you also use to multiply 39 by to find X. hope this helps!!!

6 0

3 years ago

What is 500 - 200. please answer.

Assoli18 [71]

Answer:

300

Step-by-step explanation:

4 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

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$