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Montano1993 [528]
2 years ago
14

Frankie is saving up to have spending money for her family vacation. If Frankie's current monthly net pay is $490.00 and her mon

thly expenses are $396.90 what percent of her net pay is left for savings?
19%
23%
67%
81%
Mathematics
1 answer:
grigory [225]2 years ago
8 0
We’ll first find how much she has left as her savings
$490-$396.90= $93.70

Now we’ll go for cross multiplication
100% $490
? $93.70

? = 100 x 93.7/490
? = 19.1 %

So the answer is 19.1% which rounded to the nearest whole number is 19%. The answer is the first option.
Hope it helped you


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Hoochie [10]

Answer:

40.265

Step-by-step explanation:

Divide 5 by 3 and continue dividing or mulitplying the numbers given

numbers on the top row:  Mulityply

numbers on the bottom row: Divide

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What is the value of x<br><br> Triangle with sides 10 6 x-6
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Given that the triangle is a right angled triangle, to solve for x we shall use the Pythagorean theorem given by:
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3 years ago
Sally buys a handbag. there was a discount of 30%. if sally paid, $17.50, what was the original price?
andrew11 [14]
The original price was $25. 

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Second, find out what 30% of $25 is. This is the amount of the sale discount. This is always found by mulitplying 0.300 by the item's cost $25, like this:

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3 years ago
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fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

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c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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In this question, we have that:

\mu = 197.5, \sigma = 8.3

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This is 1 subtracted by the pvalue of Z when X = 210.9. So

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b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

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c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

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3 years ago
What does this equal<br> 2 [30/5] =
lorasvet [3.4K]
The correct answer is 12
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3 years ago
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