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stira [4]
3 years ago
8

22 points !!! multiply. express your answer as a fraction simplest form. 8.33 x 144

Mathematics
2 answers:
dem82 [27]3 years ago
8 0

Answer:

99 24/25

Step-by-step explanation:

8.33*√144=

8.33*12=99 24/25 or 100 with 8.33 bar

ryzh [129]3 years ago
3 0

Answer:

100

Step-by-step explanation:

Let's convert both numbers into fraction form.

let x=8.33 bar

x=8.33 bar

10x=83.33 bar

9x=75

x=25/3

\sqrt{144}=12

\frac{25}{3}*\frac{12}{1}=25*4=100

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Please help me!!!<br><br> Hhhhhhhhhhhhhh
vekshin1

Answer:

C: -|x| + 3

Step-by-step explanation:

From the graph, we see that when y = 2, x is either +1 or -1

Also,when y = 1, x = +2 or -2

Thus,we can say that;

y = (-x) + 3 or -(x - 3)

So, we can write this in absolute value form as; y = -|x| + 3

4 0
3 years ago
What is 11 feet and 8 inches plus 2 feet and 9 inches
Agata [3.3K]

Answer:

14 feet and 5 inches

Step-by-step explanation:

11 feet + 2 feet= 13 feet

9 inches + 8 inches = 17 inches

17 inches = 1 foot and 5 inches

13 + 1 = 14 feet

5 inches + 0 inches = 5 inches

7 0
3 years ago
A researcher is interested in the lengths of brook trout, which are known to be approximately Normally distributed with mean 80
aliya0001 [1]

Answer:

P(X

And we can find this probability with the normal standard distirbution or excel and we got:

P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:

X \sim N(80,5)  

Where \mu=80 and \sigma=5

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(X

And we can find this probability with the normal standard distirbution or excel and we got:

P(z

6 0
3 years ago
10 - 2x, when x = 3<br>​
Nesterboy [21]

Answer:

4

Step-by-step explanation:

Plug in 3 as x in the expression:

10 - 2x

10 - 2(3)

10 - 6

= 4

4 0
3 years ago
Read 2 more answers
Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 1
Yuri [45]

Answer:

90% confidence interval for the population mean test score is [95.40 , 106.59]

Step-by-step explanation:

We are given that the population mean score for mathematics test is 115. It is known that the population standard deviation of test scores is 17.

Also, a random sample of 25 students take the exam. The mean score for this group is 101.

The, pivotal quantity for 90% confidence interval for the population mean test score is given by;

        P.Q. = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean = 101

              \sigma = population standard deviation

              n = sample size = 25

So, 90% confidence interval for the population mean test score, \mu is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.6449) = 0.90

P(-1.6449 * \frac{\sigma}{\sqrt{n} } < {Xbar-\mu} < 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

P(X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } < \mu < X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

90% confidence interval for \mu = [ X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } , X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ]

                                                  = [ 101 - 1.6449 * \frac{17}{\sqrt{25} } , 10 + 1.6449 * \frac{17}{\sqrt{25} } ]

                                                  = [95.40 , 106.59]

Therefore, 90% confidence interval for the population mean test score is [95.40 , 106.59] .

7 0
3 years ago
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