Ask question

Ask question

All categories

2 years ago

9

Lol plz help I need this answer ASAP

2 answers:

Tanya [424]2 years ago

5 0

Answer:

I believe the answer is ten.

Step-by-step explanation:

the line to CA is the same length as AE so you would double it to get ten for CD.

hope this helps!

hram777 [196]2 years ago

5 0

10 centimeters

because it's the diameter of this circle

You might be interested in

The price of a digital camera sold in stores is $129. It is sold on eBay for $65. What is the percent decrease from the store's

8090 [49]

What i did is multiply 65 times 100, and divided by 129,
(65)(100)= 6500
6500/129= 50.3

then to check i did multiply 50.3 by 129, and divided by 100
(50)(129)= 6450
6450/100= 64.50

I don't think its 50% or either 51%, because 50% gives something less than 65, and 51% gives more than 65. So maybe its in the middle.

to get the right percentage do 50.1 or 5.6 times 129, and then divide by 100

6 0

3 years ago

jessica is selling books during the summer to earn money for college. She earns a commission on each sale but has to pay for her

wariber [46]

E(x) = 10x - 30

the y int is -30........that means her expenses for the week was $ 30

5 0

3 years ago

Read 2 more answers

Are you good in math?

Karo-lina-s [1.5K]

Not really, is this a rhetorical question? lol

3 0

3 years ago

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

1 year ago

liubo4ka [24]

Answer:

78.5÷3.14=25

56.272×41.2=2318.4064

429×338×712=103241424

447÷513=0.8713450292

Step-by-step explanation:

hope it will help!

8 0

2 years ago