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2 years ago

9. 1/2 (16 + 4 + 4) + (6 X 4 = 8) I do not no how to do this

Mathematics

2 answers:

kap26 [50]2 years ago

5 0

I think it might be 456

lilavasa [31]2 years ago

5 0

1/2(24)+(16) ( i assume that the equal sign was meant to be subtraction. )

12+16= 28. that’s a guess but only works if the equal sign was meant to be a subtraction symbol

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A wildlife preserve features a crocodile with a documented age of 24 years, believed to be one of the oldest crocodiles in capti

zysi [14]

Answer:

0.0104

Step-by-step explanation:

Given that, the lifespans of crocodiles have an approximately Normal distribution, with a mean of 18 years and a standard deviation of 2.6 years.

So, mean, $\mu=18$

and standard deviation, $\sigma=2.6$

The z-score, for any arbitrary life span of x year, is given by

$z=\frac{x-\mu}{\sigma}$

By using the given values, we have

$z=\frac{x-18}{2.6} \cdots(i)$

The z-score for x=24, by using equation (i), is

$z=\frac{24-18}{2.6}=\frac{6}{2.6}=2.31$

From the table, the area to the left side of z=2.31 =0.98956

But, the proportion of crocodiles have lifespans of at least 24 years

= Area to the right side of the z=2.31

=1- (Area to the left side of the z=2.31)

=1-0.98956

=0.01044

So, the proportion of crocodiles that have lifespans of at least 24 years is 0.0104.

Hence, option (a) is correct.

5 0

2 years ago

Read 2 more answers

Given that X ~ N(120, 35). We survey samples of 25 and are interested in the distribution of X-bar. Find the z-score associated

lubasha [3.4K]

Answer:

Option c -2.8571

Step-by-step explanation:

z-score are calculated as

$z=\frac{xbar-mean}{\frac{S.D}{\sqrt{n} } }$ .

We are given that mean=120 and standard deviation=S.D=35 as X~N(120,35).

Sample size=n=25.

We have to find z-score for x-bar=100. So,

z-score=[100-120]/[35/√25]

z-score=[-20]/[35/5]

z-score=-20/7

z-score=-2.8571.

Thus, the z-score associated with x-bar = 100 is -2.8571.

8 0

3 years ago

The grade you make on your exam varies directly with the number of correct answers you get on the exam.Answering 15 questions co

yuradex [85]

<u>Answer:</u>

The grade you make on your exam varies directly with the number of correct answers. The constant of variation is 5

<u>Solution:</u>

Given, The grade you make on your exam varies directly with the number of correct answers you get on the exam.

Answering 15 questions correctly will give you a grade of 75 what is the.

We have to find what is the Constant of variation.

Now, according to the given information, grade  number of correct answer

Then, grade = c x number of correct answers, where c is constant of variation.

Now, substitute grade = 75 and number of correct answers = 15

$\text { Then, } 75=c \times 15 \rightarrow c=\frac{75}{15}=5$

Hence, the constant of variation is 5

5 0

2 years ago

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

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

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

2 years ago

BaLLatris [955]

Answer:

160 is the answer please tell me if im wrong

5 0

2 years ago