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3 years ago

For my little cousin:

Mathematics

2 answers:

alexira [117]3 years ago

6 0

Answer:

Step-by-step explanation:

Mars2501 [29]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is c-(-3)=15 find the number for C

Allisa [31]

Hey, your answer is C = 12
:)

6 0

3 years ago

Read 2 more answers

18 years after purchasing shares in a mutual fund for $6300, the shares are sold for $11200 what is the total return and the ann

Grace [21]

Answer:

A = 11,600

P = 6,000

Total return = 11600-6000/6000 X 100%= 93.3%= (11600/6000) (1/17)– 1= 0.03954 * 100%= 3.954= 4.0%.

Step-by-step explanation:

5 0

4 years ago

Jose is training for a marathon.If he ran 160.2 miles in january, 163.2 miles in February, 157.8 miles in march, and 159 miles i

SVETLANKA909090 [29]

Answer: I think it's 960.

Step-by-step explanation:

Add up all of the numbers:

(160.2+163.2+157.8+159) = 640.2

Find the average by diving 640.2 by how many numbers there are:

640.2 ÷ 4 = 160.05

Round that average^^:

160.05 >>> 160

Multiply by how many numbers there will be:

In this case, your job is to estimate how far Jose ran in 6 months, not just those 4. So instead of multiplying the average (160) by 4, you'd multiply it by 6.

160 x 6= 960

3 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

bulgar [2K]

Answer:

a) 20.61%

b) 21.82%

c) 42.36%

d) 4 withdrawals

Step-by-step explanation:

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

$\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}$

We are looking for

P(0;20)+P(0;1)+P(0;2) =

$\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=$

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

Here we want P(20;4)

$\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%$

Here we need

$\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)$

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

$\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%$

For a binomial distribution the expectance of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = 4 withdrawals.

3 0

3 years ago

Assessment Practice 11. The rectangle has a perimeter of 86 yards. Which is its area?

Semenov [28]

Answer:

460

Step-by-step explanation:

23+23=46. 20+20=40. 46+40=86. 23×20= 460

8 0

2 years ago