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katen-ka-za [31]

2 years ago

8

Zoe wrote 3/5 of her essay in 30 minutes. Assuming she wrote the paper at a constant rate the entire time, which expression can

be used to determine the fraction of her essay she wrote each minute?

1 answer:

LUCKY_DIMON [66]2 years ago

3 0

Answer: The answer is (B)

Step-by-step explanation:

It just took the unit test

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There are 5 marbles in a bag: 4 are blue and 1 is red. What is the probability that a blue marble gets picked?

charle [14.2K]

Answer:

the answer is 80%

Step-by-step explanation:

4/5 = 80%

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2 years ago

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A master student is planning to take a qualifying exam in the coming summer. She has three chances. The first one is in June. If

Soloha48 [4]

Answer:

Step-by-step explanation:

Given

$P_1=$ Probability of clearing exam in 1 st attempt=0.5

$P_2=$ Probability of clearing exam in 2 nd attempt=0.7

$P_3=$ Probability of clearing exam in 3 rd attempt=0.8

Probability that she passes the exam $=P(1 st\ attempt)+P(1\ fail)\cdot P(2\ pass)+P(1\ fail)\cdot P(2\ fail)\cdot P(3 rd pass)$

$P=0.5+0.5\times 0.7+0.5\times 0.3\times 0.8=0.97$

(b)P(Pass qualification on 2nd try|passes qualification)

P(Pass qualification on 2nd try|passes qualification) $=\frac{P(fail\ on\ 1 )\cdot P(pass\ in\ 2nd)}{P(passes)}$

P(Pass qualification on 2nd try|passes qualification) $=\frac{0.5\times 0.7}{0.97}=0.3608$

5 0

2 years ago

Complete the table. Round to the nearest cent if necessary.

murzikaleks [220]

Answer: 99.07$
Steps: First turn 32% into a decimal by moving the decimal right 2 places. You get 0.32. Then multiply 75.05$ by your decimal. You get 24.016. Add that to 75.05 and round your cent up, getting 99.07$ Sorry if i am incorrect this is my first answer. Hope I helped :)

6 0

2 years ago

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

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

Formula:

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

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

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2 years ago

HELP QUICK PLEASE <br><br> 12(20 + b)

sweet-ann [11.9K]

Answer:

b+9=20-12

Step-by-step explanation:

Hopefully this helped, if not HMU and I will try my best to get you a better answer!

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2 years ago

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