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leonid [27]
3 years ago
14

7. Given Rhombus FISH. find angle 1

Mathematics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

please give a picture of Rhombus FISH so i can help

Step-by-step explanation:

Ill answer once i see the picture

:D

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Jamie purchased a DVD that was on sale for 15% off. The sales tax in her county is 5%. Let y represent the original price of the
damaskus [11]

Answer: Option 'D' is correct.

Step-by-step explanation:

Since we have given that

Sale off = 15%

Rate of sales tax = 5%

Let y be the original price of the DVD.

Amount of sales tax would be

\dfrac{5}{100}\times y\\\\=0.05y

Amount after tax would be

y + 0.05 y = 1.05y

Now, there is off of 15%, so final cost of the DVD would be

\dfrac{100-15}{100}\times 1.05y\\\\=\dfrac{85}{100}\times 1.05y\\\\=0.85(1.05y)

Hence, Option 'D' is correct.

4 0
4 years ago
Read 2 more answers
Simplify.
vovangra [49]
Pretty sure it is A. x4+10x2+48
Because:
If you multiply x^2 by X^2= x4
if you multiply 5^x by 2^x- 5*2=10 and x*x= x2 so it makes 10x2 
and then 8*6=46 
so if you put that all together then it is x4+10x2+48


5 0
3 years ago
Read 2 more answers
What is the GCF (Greatest Common Factor) of (18,176)
puteri [66]

Answer:

2

Step-by-step explanation:

18 =   2 * 3 * 3

176 = 2 * 2* 2 * 2 * 11

GCF = 2

6 0
3 years ago
The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontradit
IceJOKER [234]

Answer:

94.52% probability  that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be 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}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3.5, \sigma = 0.5, n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05

What is the probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42?

This is 1 subtracted by the pvalue of Z when X = 3.42.

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

Z = \frac{3.42 - 3.5}{0.05}

Z = -1.6

Z = -1.6 has a pvalue of 0.0548.

So there is a 1-0.0548 = 0.9452 = 94.52% probability  that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.

6 0
3 years ago
A rectangle has a perimeter of 84cm and a length 35cm what is its width
aleksandrvk [35]
P = 2(L + W)
P = 84
L = 35

now we sub
84 = 2(35 + W)
84 = 70 + 2W
84 - 70 = 2W
14 = 2W
14/2 = W
7 = W....so the width is 7 cm
5 0
3 years ago
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