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Alex Ar [27]
3 years ago
8

The weights of healthy adult male Labrador Retrievers are approximately Normally distributed with a mean of 77 pounds and a stan

dard deviation of 6 pounds
A)Make an accurate sketch of the weights of Labrador Retrievers with the horizontal axis marked

B)What weight would represent the 25th percentile?

C)What weight would represent the 75th percentile?

D) What is the interquartile range of male Labrador weights?

E)What proportion of male adult Labrador retrievers weigh above 85 pounds?

F) What proportion of male adult Labrador retrievers weigh between 60 and 80 pounds?

G) What proportion of male adult Labrador retrievers weigh less than 65 pounds?
Mathematics
1 answer:
gavmur [86]3 years ago
6 0

Answer:the photo is for part A.

B) the weight would be Invnorm(.25,77,6)=72.95

C)the weight would be 81.04 but this is the work invnorm(.75,77,6)=81.04

D)IQR=Q3-Q1

81.04-72.95=8.09

E)9.12 is the answer the work normalcdf(85,1000,77,6)=.0912

F).6892 or 68.92% are the answers the work normalcdf(60,80,77,6)=.6892

G).0227 or 2.27% is the answer and the work normalcdf(0,65,77,6)=.0227

Step-by-step explanation:

You might be interested in
What is 100/7 divided by 4 1/3 simplified if you answer thank you and happy holidays
SVETLANKA909090 [29]

Answer: 300/91

Step-by-step explanation:

1. Convert 4 1/3 to an improper fraction

13/3

2. Divide, using the keep, change, flip method to divide fractions

- Keep the first fraction the same (100/7)

- Change the division sign to a multiplication sign

- Flip the second fraction (13/3 -> 3/13)

3. Plug-in and evaluate

100/7 * 3/13

300/91

If you are looking for the solution as a mixed number, it is 3 27/91

4. Simplify, if applicable. This fraction cannot be simplified any further.

7 0
3 years ago
While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between
aleksandr82 [10.1K]

Answer:

The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

Step-by-step explanation:

Let the random variable <em>X</em> denote the water depths.

As the variable water depths is continuous variable, the random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 2.00 m and <em>b</em> = 7.00 m.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:

P(2.25

                               =\frac{1}{5.00}\int\limits^{5.00}_{2.25} {1} \, dx\\\\=0.20\times [x]^{5.00}_{2.25} \\\\=0.20\times (5.00-2.25)\\\\=0.55

Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

6 0
3 years ago
Amanda uses 1/3 of a cup of milk each time she makes a batch of pancakes. How manybbatches can she make if she only has 11/22 of
kow [346]

Answer:

<u>Amanda can make 1.5 or 3/2 of batches of pancakes with 11/22 of a cup of milk.</u>

Step-by-step explanation:

1. Let's review the information given to us for answering the question correctly:

Amount of milk Amanda use for a batch of pancakes = 1/3 cup

2. How many batches can she make if she only has 11/22 of a cup of milk left?

Let's simplify 11/22, dividing by 11 the numerator and the denominator,

11/22 = 1/2

Now, using the Rule of Three Simple, we can calculate the answer to this question, this way:

Number of batches = (1/2 * 1)/1/3

Number of batches = 1/2 / 1/3

Number of batches = 1/2 * 3/1 (Taking the reciprocal)

Number of batches = 3/2 = 1.5

<u>Amanda can make 1.5 or 3/2 of batches of pancakes with 11/22 of a cup of milk.</u>

4 0
2 years ago
A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is \frac{1}{13}.

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is 4\times 51!.

The probability that the last card dealt is an ace is

P=\frac{4\times 51!}{52!}

P=\frac{4\times 51!}{52\times 51!}

P=\frac{4}{52}

P=\frac{1}{13}

Therefore, the probability that the last card dealt is an ace is \frac{1}{13}.

4 0
3 years ago
Which graph decreases, crosses the y-axis at (0,-7), and then remains constant?
sergey [27]

Answer:

A.Graph B

Step-by-step explanation:

it decreases to a y of (0,-7) and then stays constant at -7

Hope this was helpful

7 0
3 years ago
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