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Rina8888 [55]
3 years ago
12

1) Mrs. Merson is selling her car. Her research shows that the car has a current value of $5,500,

Mathematics
1 answer:
SVEN [57.7K]3 years ago
0 0

The right answer is Option B:  \frac{5500}{p}=\frac{55}{100}

Step-by-step explanation:

Given,

Current value of car = $5500

This is 55% of the original value.

Let,

p be the original price of car.

Therefore,

55% of p = 5500

\frac{55}{100}p=5500

Dividing both sides by p

\frac{55}{100}*\frac{p}{p}=\frac{5500}{p}\\\frac{55}{100}=\frac{5500}{p}

The equation \frac{5500}{p}=\frac{55}{100} can be used to find the price Mrs. Merson paid for the car.

The right answer is Option B:  \frac{5500}{p}=\frac{55}{100}

Keywords: percentage, division

Learn more about percentages at:

  • brainly.com/question/13062539
  • brainly.com/question/13076219

#LearnwithBrainly

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What equation best represents the following sentence?
Tems11 [23]

Answer:

2n + 1 to find the first odd number.

Step-by-step explanation:

that is it

8 0
3 years ago
3. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population p
damaskus [11]

Answer:

A)sample proportion = 0.17,  the sampling distribution of p can be calculated/approximated with normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

B) 0.869

C)0.9668

Step-by-step explanation:

A) p ( proportion of population that spends more than $100 per week) = 0.17

sample size (n)= 800

the sample proportion of p = 0.17

standard error of p = \sqrt{\frac{p(1-p)}{n} } = 0.013281

the sampling distribution of p can be calculated/approximated with

normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

B) probability that the sample proportion will be +-0.02 of the population proportion

= p (0.17 - 0.02 ≤ P ≤ 0.17 + 0.02 ) = p( 0.15 ≤ P ≤ 0.19)

z value corresponding to P

Z = \frac{P - p}{standard deviation}

at P = 0.15

Z =  (0.15 - 0.17) / 0.013281 = = -1.51

at P = 0.19

z = ( 0.19 - 0.17) / 0.013281 = 1.51

therefore the required probability will be

p( -1.5 ≤ z ≤ 1.5 ) = p(z ≤ 1.51 ) - p(z ≤ -1.51 )

                           = 0.9345 - 0.0655 = 0.869

C) for a sample (n ) = 1600

standard deviation/ error = 0.009391 (applying the equation for calculating standard error as seen in part A above)

therefore the required probability after applying

z = \frac{P-p}{standard deviation} at p = 0.15 and p = 0.19

p ( -2.13 ≤ z ≤ 2.13 ) = p( z ≤ 2.13 ) - p( z ≤ -2.13 )

                               = 0.9834 - 0.0166 = 0.9668

7 0
3 years ago
Read 2 more answers
The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in
Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 64 inches

           \sigma = standard deviation = 4 inches

           n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{63-64}{\frac{4}{\sqrt{4} } } ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

                                                        = 1 - 0.69146 = 0.30854

Hence, it is 30.85%  likely that average height would be shorter than 63 inches.

7 0
3 years ago
22
dalvyx [7]

The first digit may be 1 or 8.

The second digit may be 1, 2, 4, 8.

The third digit may be 1, 3, 5, 7, 9.

There are 2\times4\times5=40\:variants

The probability that this is the correct code is \frac{1}{40}

8 0
2 years ago
What is the probability of getting heads and a b, c, and d
sp2606 [1]
I mean probably of getting head is a half if that's what u mean but what is a,b,c,d
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