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Hitman42 [59]
3 years ago
10

PLS HELP! (THE QUESTION IS A PICTURE CLICK ON THE QUESTION FIRST)

Mathematics
2 answers:
Natalija [7]3 years ago
7 0

Answer:

Step-by-step explanation:

B.28.1

KengaRu [80]3 years ago
4 0

Answer:

It's BBBBBBBBBBBBBBBBB

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Michelle is trying to buy a car that costs $35,550, and she has saved up $20,700. She borrowed the remaining amount that she nee
Dimas [21]

Answer: $ 4,515

Step-by-step explanation:

\\The cost of the car she wanted to buy = $35,550

\\Her savings = $20,700

\\That means she needs( $35,550 - $20,700) more in order for her to buy the car.

\\Amount borrowed from the finance company  = $35,550 - $20,700

\\= $ 14,850

\\Since she now owes the financial company a total of $19,365, then the interest accrued by the loan she borrowed is given by:

\\$19,365 - $ 14,850

\\= $4,515

7 0
3 years ago
The average weight of full-grown beef cows is 1,470 pounds with a standard deviation of 230 pounds. if the weights are normally
creativ13 [48]

Answer: The percentile is 89

Step-by-step explanation:

This question can be solved using concept for t tables

In a normal distribution the curve. \mu= 1470 ,  \sigma = 230, x=1750

The relationship between z score, mean and standard deviation is given by

x = \mu+z\sigma

So the z value according to this is given by the formula

1750=1470+z(230)\\ z=\frac{280}{230} \\\\z = 1.217

From the z table we can infer that p value for z=+1.217 is 88.82

So 1750 is 89th percentile

To learn more about statistics, visit brainly.com/question/26352252

#SPJ4

8 0
1 year ago
Write an inequality to model the solutions shown on the number line below.
DedPeter [7]

Answer:

X > 6

Step-by-step explanation:

5 0
3 years ago
Find the minimum sample size needed to estimate the percentage of Democrats who have a sibling. Use a 0.1 margin of error, use a
Brums [2.3K]

Answer:

The minimum sample size is  n  =135

Step-by-step explanation:

From the question we are told that

   The confidence interval is ( lower \ limit  = \  0.44,\ \ \   upper \ limit  = \ 0.51)

    The margin of error is  E =  0.1

   

Generally the sample  proportion can be mathematically evaluated as

     \r p = \frac{ upper \ limit  + lower \ limit }{2}

    \r p = \frac{ 0.51 + 0.44}{2}

    \r p = 0.475

Given that the confidence level is  98% then the level of significance can be mathematically evaluated as

         \alpha = 100 -  98

        \alpha = 2\%

        \alpha =0.02

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table  

   The value is

        Z_{\frac{\alpha }{2} } =  2.33

Generally the minimum sample size is evaluated as

      n  =[ \frac { Z_{\frac{\alpha }{2} }}{E} ]^2 *  \r p (1- \r p )

     n  =[ \frac { 2.33}{0.1} ]^2 *  0.475(1- 0.475 )

     n  =135

6 0
3 years ago
The Greek mathematician Eratosthenes (ca. 276-195 BC) measured the circumference of the earthfrom the following observations. He
denpristay [2]

Answer:

Radius of the Earth is 3978.8 Miles

Circumference of the Earth is 25000 Miles

Step-by-step explanation:

The angle of the sun shone at an angle of 7.2° to the zenith

This means that the angle of the sector of the circle is 7.2° (θ)

S = Length of the sector of the circle = 500 miles

r = radius of earth

Converting 7.2° to radians

\theta =7.2\frac{\pi}{180}

S=r\theta\\\Rightarrow r=\frac{S}{\theta}\\\Rightarrow r=\frac{500}{7.2\frac{\pi}{180}}\\\Rightarrow r=3978.8\ Miles

∴ Radius of the Earth is 3978.8 Miles

C=2\pi r\\\Rightarrow C=2\times \pi \frac{500}{7.2\frac{\pi}{180}}\\\Rightarrow C=25000\ Miles

∴ Circumference of the Earth is 25000 Miles

5 0
4 years ago
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