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Mashcka [7]
3 years ago
5

Refer to the data set below​ (body mass index of​ men) and determine whether the requirement of a normal distribution is satisfi

ed. Assume that this requirement is loose in the sense that the population distribution need not be exactly​ normal, but it must be a distribution that is basically symmetric with only one mode.
84 79 82 74 78

Is the requirement of a normal distribution​ satisfied?
Mathematics
1 answer:
AnnyKZ [126]3 years ago
4 0

Answer:

Yes . The requirement of a normal distribution​ are satisfied.

Step-by-step explanation

The points of the data in the normal quantile plot lie close to a straight line.

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Verify the identity<br><br> cos quantity x plus pi divided by two = -sin x
In-s [12.5K]

Answer:

Identity is verified.

Step-by-step explanation:

We have to verify the identity cos(x+\frac{\pi }{2}) = (- sinx)

To prove any identity we always prove one side(either left hand side or right hand side) of the equation equal to the other side.

In this identity we take the left hand side first

cos(x+\frac{\pi}{2})

=cosx\times cos(\pi/2)-sinx\times sin(\pi/2))  (as we know cos(a+b) = cosa×cosb-sina×sinb)

= cosx\times0-sinx\times1

= 0-sinx

= - sinx ( Right hand side)

Hence identity is proved.

6 0
3 years ago
A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choo
Keith_Richards [23]

Answer:

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

 

Step-by-step explanation:

 Given the data in the question;

To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

Alternative hypothesis H₁ : = < (75%) < 0.75

now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

so, the number of times p"<0.64 in 150 simulations is 5

Hence; P(p" < 0.64 ) = 5 / 150 = 0.0333

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

7 0
3 years ago
Clare is painting some doors that are all the same size. She used 2 liters of paint to cover 1 3/5 doors. How many liters of pai
barxatty [35]

Answer:

1\frac{1}{4}\ liters

Step-by-step explanation:

step 1

Convert mixed number to an improper fraction

1\frac{3}{5}=\frac{1*5+3}{5}=\frac{8}{5}

step 2

Using proportion

\frac{2}{(8/5)}\frac{liters}{doors} =\frac{x}{1}\frac{liters}{door}\\ \\x=2/(8/5)\\ \\x=10/8\ liters

step 3

Convert to mixed number

\frac{10}{8}\ liters=\frac{8}{8}+\frac{2}{8}=1\frac{1}{4}\ liters

8 0
3 years ago
Please do provide a correct answer because other ways I will report/block you. Please be honest and help me, please.
Irina-Kira [14]

Answer:

incomplete

Step-by-step explanation:

please send the full information, I can't see students belonging to acers

4 0
2 years ago
A fruit seller packed some apples into 15 small cartons and 20 big cartons. Each big carton contained 15 more apples than each s
ehidna [41]

Answer:

720 apples

Step-by-step explanation:

Let A be the number of apples in the small pack.

QTY for Small Cartons is 15A,

QTY for Big Cartons is 20(A+15) = 20A+300

From small cartons makes 25% or 1/4 of total, means the big carton makes 75% of total, or 3/4 of total. This also means the qty in big carton is 3 times of small carton

20A+300 = 3X (15A)= 45A

300 = 25A

A=12

Altogether = 15A+(20A+300) = 720 Apples

7 0
2 years ago
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