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Vlada [557]
2 years ago
12

I need help pleases I wil give brainyless

Mathematics
2 answers:
erica [24]2 years ago
3 0

Answer: Let's start by writing down coordinates of all points:

A(0,0,0)

B(0,5,0)

C(3,5,0)

D(3,0,0)

E(3,0,4)

F(0,0,4)

G(0,5,4)

H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate  changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)

Reflecting

A(0,0,0)

B(0,5,0)

Reflecting

B(0,-5,0)

C(3,5,0)

Reflecting

C(3,-5,0)

D(3,0,0)

Reflecting

D(3,0,0)

b.)

A(0,0,0)

Moving

A(-2,-3,1)

B(0,-5,0)

Moving

B(-2,-8,1)

C(3,-5,0)

Moving

C(1,-8,1)

D(3,0,0)

Moving

D(1,-3,1)

Hope this helps plz mark brainliest

Step-by-step explanation:

mestny [16]2 years ago
3 0

Answer:

It seems like this one is more of an estimate based off the graph so I would go with either 1 or 0 but most likely 1

Step-by-step explanation:

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Solve for y <br> 7+4y=7-5y+54
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Answer:

Step-by-step explanation:

7+4y=7-5y+54

Collecting like terms

4y + 5y = 7 - 7 + 54

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Find the mid point of each line segment​
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Answer:

The midpoint is (2, -2) :)

Step-by-step explanation:

Midpoint formula time!

(3 + 1) / 2     +        (0 - 4) / 2

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3 years ago
Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce
LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 ​kg.

92.82% of the sample means are between 75 kg and 83 ​kg.

Step-by-step explanation:

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}}.

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:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that \mu = 79, \sigma = 22.

What percentage of individual adult females have weights between 75 kg and 83 ​kg?

This percentage is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 75. So:

X = 83

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

Z = \frac{83 - 79}{22}

Z = 0.18

Z = 0.18 has a pvalue of 0.5714.

X = 75

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

Z = \frac{75- 79}{22}

Z = -0.18

Z = -0.18 has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 ​kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each​ sample, what percentage of the sample means are between 75 kg and 83 ​kg?

Now we use the Central Limit THeorem, when n = 100. So s = \frac{22}{\sqrt{100}} = 2.2.

X = 83

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

Z = \frac{83 - 79}{2.2}

Z = 1.8

Z = 1.8 has a pvalue of 0.9641.

X = 75

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

Z = \frac{75-79}{2.2}

Z = -1.8

Z = -1.8 has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 ​kg.

8 0
3 years ago
Which of the following points is the reflection of (8, -5) in the y-axis?
MAVERICK [17]

Answer:

D

Step-by-step explanation:

When you flip a point around the y axis the x value is inverted since the point is moved to the opposite side across the y axis.

5 0
2 years ago
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