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Makovka662 [10]
3 years ago
8

Find the surface area of the pyramid.

Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0

Answer: 56.5

Step-by-step explanation: hope this helps you!!

Butoxors [25]3 years ago
8 0
The surface area of the pyramid would be 56.5
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What does 15x+35 equal
____ [38]
<span>15x+35 = 5(3x + 7)

hope that helps</span>
7 0
3 years ago
Gary's rental costs are $135.69, and he must pay 5 percent tax, plus $14 per day for insurance for 3 days. How much will he pay
klio [65]
5% tax of 135.69
Multiply 135.69 * .05 = 6.7845
5% tax of 135.69 is 6.78
135.69+6.78=142.47

14*3=42

142.47 + 42 = 184.47

Gary will pay $184.47 in all.


Hope this helps you! (:
-Hamilton1757
4 0
3 years ago
Given that K is the centroid of EFG find GE and GI
Anastaziya [24]

Answer:

D) GE = 10 and GI = 12

Step-by-step explanation:

Given: KI = 4 and GH = 5

K is the centroid of EFG

So

KI = 1/3 (GI)

GI = 3 (KI) = 3(4) = 12

Because K is the centroid of EFG so GH = HE = 5

GE = GH + HE

GE = 5 + 5

GE = 10

Answer

GE = 10 and GI = 12

5 0
4 years ago
Read 2 more answers
Please help, will mark brainliest!!
Novay_Z [31]

Answer:

$12000

Step-by-step explanation:

P=CRT

C=cost

R=rate

T=time

P=3000 x 8 x 50/100= $12000

3 0
2 years ago
Consider the probability that at least 91 out of 155 students will pass their college placement exams. Assume the probability th
Hatshy [7]

Answer:

0.5616 = 56.16% probability that at least 91 out of 155 students will pass their college placement exams.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 155, p = 0.59

So

\mu = E(X) = np = 155*0.59 = 91.45

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{155*0.59*0.41} = 6.12

Probability that at least 91 out of 155 students will pass their college placement exams.

Using continuity correction, this is P(X \geq 91 - 0.5) = P(X \geq 90.5), which is 1 subtracted by the pvalue of Z when X = 90.5. So

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

Z = \frac{90.5 - 91.45}{6.12}

Z = -0.155

Z = -0.155 has a pvalue of 0.4384

1 - 0.4384 = 0.5616

0.5616 = 56.16% probability that at least 91 out of 155 students will pass their college placement exams.

4 0
3 years ago
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