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anyanavicka [17]

3 years ago

12

a toy company need to package balls with a 2in radius in sets of 2. they have two options: a square prism or a cylinder. The dim

ensions are shown below

2 answers:

ser-zykov [4K]3 years ago

8 0

Answer:

a.the cylinder because it has about 27.5 in.3 less wasted space than the prism

Step-by-step explanation:

got it right on edge

Brut [27]3 years ago

6 0

Answer:

Step-by-step explanation:

square prism

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When comparing two data sets, Kimberly uses the median and IQR after determining that the mean and MAD do not accurately represe

77julia77 [94]

Answer:

C

Step-by-step explanation:

3 0

3 years ago

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vazorg [7]

Answer:

rice --- no of people

1 ---- 3

2 ---- 6

4 ---- 12

12 --- 36

43 --- 129

Step-by-step explanation:

This is in direct proportion. The constant value of it is 3 so you have to multiply the number of cups into three to get the number of people it can feed

3 0

3 years ago

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67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

$P(X=r) = nCr p^r q^{n-r}$

where $nCr = \frac{n!}{(n-r)! r!}$

a) Probability that exactly 29 of them are spayed or neutered

$P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074$

b) Probability that at most 33 of them are spayed or neutered

$P(X \leq 33) =1 - P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66$

c) Probability that at least 30 of them are spayed or neutered

$P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79$

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

$P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574$

8 0

3 years ago

Write a two column proof.

lbvjy [14]

Answer:

this should be correct.

7 0

3 years ago

Sally is counting the people and animals that live on her farm. She, her brother, and her parents together have 18 cows, 1 horse

Bas_tet [7]

Using it's concept, it is found that there is a:

0.6207 = 62.07% probability of selecting a cow.

0.0345 = 3.45% probability of selecting a horse.

0.3448 = 34.48% probability of selecting a goat.

0% probability of selecting a chicken.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, there is a total of 18 + 1 + 10 + 0 = 29 animals, hence:

18/29 = 0.6207.

0.6207 = 62.07% probability of selecting a cow.

1/29 = 0.0345.

0.0345 = 3.45% probability of selecting a horse.

10/29 = 0.3448.

0.3448 = 34.48% probability of selecting a goat.

0/29 = 0.

0% probability of selecting a chicken.

More can be learned about the probability concept at brainly.com/question/15536019

3 0

2 years ago