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3 years ago

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Cara uses nylon cord to make necklaces. Each necklace requires 18.75 inches of cord. Cara has 281.25 inches of the cord on hand.

If she uses all of it, how many necklaces can Cara make?

1 answer:

Juli2301 [7.4K]3 years ago

3 0

Answer:

Cara can make 15 nylon cord necklaces exactly.

Step-by-step explanation:

This looks like a simple division problem, just divide 281.25 by 18.75 and you should get 15.

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What is the unit rate for 15 apples eaten in 5 minutes for 15 apples eaten in 5 minutes

USPshnik [31]

The unit rate is 3 apples per minute

Step-by-step explanation:

Unit rate will be calculated by dividing the total number of apples by the time

Given

Total number of apples = 15

Time = 5 min

So, The unit rate will be:

$Unit\ rate=\frac{Apples}{Time}\\=\frac{15}{5}\\=3\ apples\ per\ min$

Hence,

The unit rate is 3 apples per minute

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3 years ago

A study indicates that 62% of students have have a laptop. You randomly sample 8 students. Find the probability that between 4 a

Scrat [10]

Answer:

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a laptop, or they do not. The probability of a student having a laptop is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

A study indicates that 62% of students have have a laptop.

This means that $n = 0.62$

You randomly sample 8 students.

This means that $n = 8$

Find the probability that between 4 and 6 (including endpoints) have a laptop.

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157$

$P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815$

$P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297$

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269$

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

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3 years ago

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Read 2 more answers

Morgan is walking her dog on an 8-meter-long leash. She is currently 500 meters from her house, so the maximum and minimum dista

Gnom [1K]

Given:
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minimum distance of the dog from the house 500 - 8 = 492 meters.
maximum distance of the dog from the house 500 + 8 = 508 meters.

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