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astra-53 [7]
3 years ago
3

if it can be assumed that the population is normal, then what is the probability that one man sampled from this population has a

weight between 72kg and 88kg
Mathematics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

The probability that one man sampled from this population has a weight between 72kg and 88kg is 0.6826.

Step-by-step explanation:

The complete question has the data of mean = 80 kg and standard deviation = 8kg

We have to find the probability between 72 kg and 88 kg

Since it is a normal distribution

(x`- u1 / σ/ √n) < Z >( x`- u2 / σ/ √n)

P (72 <x>88) = P ( 72-80/8/√1) <Z > ( 88-80/8/√1)

= P (-1<Z> 1) = 1- P (Z<1) - P (Z<-1)

= 1- 0.8413- (- 0.8413)= 1- 1.6826= 0.6826

So the probability that one man sampled from this population has a weight between 72kg and 88kg is 0.6826.

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Which size ice cream dessert gives you the best price per ounce? Small 6-oz for $2.49. Medium 10-oz cup for $3.49. Large 16-oz c
natulia [17]

Answer:

The large size gives the lowest price per ounce of 3.11875, therefor the best in terms of price

Step-by-step explanation:

Step 1: Determine expression for price per ounce

The expression for the total price of an ice cream dessert is as follows;

T=p×n

where;

T=total price of ice cream

p=price per ounce of ice cream

n=number of ounces of ice cream

Step 2: Convert total price to price per ounce

The expression can be rewritten as;

p=T/n

1. For small 6 ounce at $2.49, T=2.49, n=6-oz

p=2.49/6=0.415

The price per ounce of the small ice cream dessert=$0.415 per ounce

2. For the Medium 10-oz for $3.49, T=$3.49, n=10-oz

p=3.49/10=0.349

The price per ounce of the medium ice cream dessert=$0.349 per ounce

3. For the Large 16-oz for $4.99, T= $4.99, n=16

p=4.99/16=$0.311875

The price per ounce of the medium ice cream dessert=$0.311875 per ounce

4. For the super size 24-oz for $7.69, T=$7.69, n=24

p=7.69/24=0.32

The price per ounce of the super size cup is 0.32 per ounce

The large size gives the lowest price per ounce of 3.11875, therefor the best in terms of price

3 0
3 years ago
We are to estimate the confidence interval for a population proportion p, based on two samples A and B of sizes 100 and 200, res
WITCHER [35]

Answer:

The option B is True

Step-by-step explanation:

Solution

nA  < nB

Thus

1/√nA > 1/√nB

SD/√nA  > SD/√nB

So,

SE (A) > SE (B).......(1)

Zα/2 SE (A) > Zα/2 SE (B)

Now

E (A) > E(B)  this is the margin error

Since SE fro sample A is larger than SE for sample B

We have that,

The length = ZE

Therefore, option B is correct because here, we know that as the sample size increases or goes higher  the length of the confidence interval decreases

In this case, the  sample size is large for B then A.

Thus, the length of confidence interval based on A is larger than the length of confidence interval Based on B Since SE is large For A than B

7 0
3 years ago
$70 plus a 5% tax and a 15% tip
Serjik [45]

Answer:

maybe 88.5

Step-by-step explanation:

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5 0
3 years ago
Please help guys ):​
qaws [65]

Answer:

∠ B = 126°

Step-by-step explanation:

Since the lines are parallel, then

∠ A = ∠ B ( alternate angles ), thus

8x + 78 = 2x + 114 ( subtract 2x from both sides )

6x + 78 = 114 ( subtract 78 from both sides )

6x = 36 ( divide both sides by 6 )

x = 6

Thus

∠ B = 2x + 114 = 2(6) + 114 = 12 + 114 = 126°

4 0
3 years ago
What is the answer to 6x-3(2-3x)
insens350 [35]

Problem: 6x-3(2-3x)

<em>Let's solve!</em>

<em />

First: we expand

6x-(6-9x)

Then we simplify

6x-6+9x

Collect like terms

(6x+9x)-6

Simplify:

15x-6

The answer is 15x-6

5 0
3 years ago
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