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2 years ago

7

A picture on a page was reduced by 60% of its original size, and this copy was then reduced by 20%. What percent of the size of

the original picture was final copy ?

1 answer:

melamori03 [73]2 years ago

6 0

Answer:

the final copy should be 12% of the original picture.

Step-by-step explanation:

you multiply 20% by 60%

and you get 0.12 which is the same as 12%

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which statement below is represented by the correct ratio?Choose all that apply. for ever 50 shoppers 25 watermelons are sold

ASHA 777 [7]

For every 50 shoppers, 25 watermelons are sold, so 50/25

Brainliest?
I'm trying to move up a rank and it would help!

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3 years ago

Need help with this please!! The scatter plot shows the results of a survey in which 10 students were asked how many hours they

Anna11 [10]

5 students scored a 90 or above. If you look at the axis that says “test scores” just count the number of dots on the “90” line and above

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3 years ago

Read 2 more answers

When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil

Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability 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:

90% of the students passed, hence $p = 0.9$ .

The professor randomly selected 10 exams, hence $n = 10$ .

Item a:

The probability is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which:

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

$P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937$

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

$P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487$

Then:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298$

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

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

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

$P(X < 5) = 0.0001$

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

3 0

2 years ago

In a random sample of 45 newborn babies, what is the probability that 40% or fewer of the sample are boys. (Use 3 decimal places

erastova [34]

Answer:

$z= \frac{p- \mu_p}{\sigma_p}$

And the z score for 0.4 is

$z = \frac{0.4-0.4}{\sigma_p} = 0$

And then the probability desired would be:

$P(p$

Step-by-step explanation:

The normal approximation for this case is satisfied since the value for p is near to 0.5 and the sample size is large enough, and we have:

$np = 45*0.4= 18 >10$

$n(1-p) = 45*0.6= 27 >10$

For this case we can assume that the population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Where:

$\mu_{p}= \hat p = 0.4$

$\sigma_p = \sqrt{\frac{p(1-p)}(n} =\sqrt{\frac{0.4(1-0.4)}(45}= 0.0703$

And we want to find this probability:

$P(p$

And we can use the z score formula given by:

$z= \frac{p- \mu_p}{\sigma_p}$

And the z score for 0.4 is

$z = \frac{0.4-0.4}{\sigma_p} = 0$

And then the probability desired would be:

$P(p$

7 0

3 years ago

Read 2 more answers

Amy sold 338 flowers at the market today. If each bunch has 13 flowers in it, how many bunches did Amy sell?

Lady_Fox [76]

Answer:

26

Step-by-step explanation:

338 / 13 = 26

26 batches of 13 flowers are equal to 338.

8 0

2 years ago