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Vesna [10]
3 years ago
11

Adult Great Basin rattlesnakes have a mean length of 32.3 inches and a standard deviation of 5.3 inches; the length of adult Sou

thern Pacific rattlesnakes is also 32.3 inches on average, but with a standard deviation of 7.95 inches. Both species have lengths that follow a normal distribution. You select a random sample of 36 Great Basin rattlesnakes and 100 Southern Pacific rattlesnakes. Which is more likely be shorter than 29.65 inches
Mathematics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

POOOOOOOOOOP KAKAKAKA

Step-by-step explanation:

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Figure ABCD is transformed to obtain figure A′B′C′D′:
Kisachek [45]

Step-by-step explanation:

part A:

ABCD is transformed to obtain figure A′B′C′D′:

1) by reflection over x-axis, obtain the image :

A(-4,-4) B(-2,-2) C(-2, 1) D(-4, -1)

2) by translation T (7 0), obtain the image :

A'(3,-4) B'(5,-2) C'(5, 1) D'(3, -1)

part B:

the two figures are congruent.

the figures that transformed by reflection either or translation will obtain the images with the same shape and size (congruent)

6 0
3 years ago
-2x+y=9<br><br><br> I need to isolate y
anygoal [31]

Answer:

y =2x+9

Step-by-step explanation:

-2x+y=9

Add 2x to each side

-2x+2x+y=2x+9

y =2x+9

7 0
3 years ago
Read 2 more answers
50 points and brainliest for the best fastest answer
romanna [79]

Answer:

ha

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please help me due tomorrow
Tom [10]

<u>Given </u><u>:</u><u>-</u>

  • The slope of the line through points (3,y) and (4,10) is 7 .

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The value of y .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As we know that the slope of the line is difference of ordinate divided by the difference of absicca as ,

\longrightarrow m = y -10 / 3 - 4

\longrightarrow 7 (-1) = y -10

\longrightarrow -7 = y -10

\longrightarrow y = 10 -7

\longrightarrow y = 3

<u>Hence</u><u> the</u><u> required</u><u> answer</u><u> is</u><u> </u><u>3.</u>

5 0
3 years ago
Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t
krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

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}

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

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

By the Central Limit Theorem

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

Z = \frac{1476 - 1467}{13.2857}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 1458

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

Z = \frac{1458 - 1467}{13.2857}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

5 0
3 years ago
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