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Olin [163]
2 years ago
12

Find the volume and surface area of prism​

Mathematics
1 answer:
dedylja [7]2 years ago
4 0

Follow up from the attachment.

Figure 2

Surface area:319cm²

Volume:184cm²

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63 years :) Hope this helps ;D
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3 years ago
Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A. Decide in each case whether R
andre [41]

Answer:

Step-by-step explanation:

Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A.

Given that (a,b) R (c,d) if ad=bc

Or (a,b) R (c,d) if determinant

\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =0

a) Reflexive:

We have (a,b) R (a,b) because ab-ab =0 Hence reflexive

b) Symmetric

(a,b) R (c,d) gives ad-bc =0

Or da-cb =0 or cb-da =0 Hence (c,d) R(a,b). Hence symmetric

Download docx
5 0
3 years ago
In a class of 147 students, 95 are taking math (M), 73 are taking science (S), and 52 are taking both math and science. What is
Marina86 [1]
95/294 orrr 0.3231 And you just round that to the nearest 10th
5 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST
coldgirl [10]

Answer:

1)4x+4y=1, 6x−y=1

x+y=4, x−y=2

x+y=−2,x+2y=2

2)2x+y=3, 2x−8y=19

Step-by-step explanation:

hope this helps

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6 0
2 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

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:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7

This probability is 1 subtracted by the pvalue of Z when X = 204.1.

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

Z = \frac{204.1 - 201.9}{0.7}

Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0
3 years ago
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