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2 years ago

Please help I’m in the exam

Mathematics

2 answers:

Iteru [2.4K]2 years ago

6 0

20

Islamic Gavin panic panic chic Iranian icon used

iVinArrow [24]2 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

The ages (in years) of the 6 employees at a particular computer store are 44, 46, 40, 34, 29, 41

son4ous [18]

step 1
compute the average: add the values and divide by 6
Average =(44+ 46+40+34+29+41)/6=39

step 2
Compute the deviations from the average
dev: (44-39)=5,
dev: (46-39)=7
dev: (40-39)=1
dev: (34-39)=-5
dev: (29-39)=-10
dev: (41-39)=2

step 3
Square the deviations and add
sum (dev^2): 5^2+7^2+1^2+-5^2+-10^2+2^2
sum (dev^2): 25+49+1+25+100+4-----> 204

step 4
Divide step #3 by the sample size=6
(typically you divide by sample size-1 to get the sample standard deviation,
but you are assuming the 6 values are the population,
so
no need to subtract 1, from the sample size.
This result is the variance
Variance =204/6=34

step 5
Standard deviation = sqrt(variance)
standard deviation= √(34)------> 5.83

the answer is
5.83

4 0

2 years ago

I need help what is the surface area of the triangular pyramid in the figure?

mr_godi [17]

Answer:

89.4 in²

Step-by-step explanation:

I'm too lazy to explain.

8 0

2 years ago

Solve for j. 3 + 4 2 -3

xeze [42]

Answer:

Hello!

Step-by-step explanation:

How do we solve for j when there is no equation with j in it to solve for j. Hope we can help you with this when we get a equation!

3 0

3 years ago

a 34 gram sample of a substance that's used to treat thyroid disorders has a k-value of 0.137. find the substance's half-life, i

Ira Lisetskai [31]

Answer:

5.1 days

Step-by-step explanation:

Given in the question,

initial amount of substance = 34 grams

k-value = 0.137

To find the half life of this substance we will use following formula

$N(0)/2 = N(0)e^{-kt}$

here N(0) is initial amount of substance

t is time in days

Plug values in the formula

$34 /2 = 34e^{-0.137t}$

1/2 = e^{-0.137t}

Take logarithm on both sides

ln(1/2) = ln( e^{-0.137t})

ln(1/2) = -0.137t

t = ln(1/2) / -0.137

t = 5.059

t ≈ 5.1 days (nearest to tenths)

3 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago