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2 years ago

Assume that X has a normal distribution, and find the indicated probability.

Mathematics

2 answers:

nlexa [21]2 years ago

8 0

Answer:

Step-by-step explanation:

m_a_m_a [10]2 years ago

6 0

Answer:

Step-by-step explanation:

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Please answer the question quickly

Zina [86]

Answer:

Step-by-step explanation:

You can already knock out the first two because you cant have negative feet. If you line the vertex (1) with the parabola, you can see that the vertex lines up with 16. You round that up and your answer should be 20

6 0

3 years ago

A collection of 108 coins containing only quarters and nickels is worth $21 wich value could replace q on the chart?

Greeley [361]

Answer:

There would be 30 nickels and 78 quarters.

Step-by-step explanation:

To find this, start by creating a system of equations in which x is the number of nickels and y is the number of quarters. Make the first equation the total number of coins.

x + y = 108

Make the second equation based on the value.

.05x + .25y = 21

Now multiply the second equation by -4 and add together to get the y value to cancel.

x + y = 108

-.2x - y = -84

--------------------

.8x = 24

x = 30

Now use this value to find y.

x + y = 108

30 + y = 108

y = 78

5 0

3 years ago

Read 2 more answers

Please help asap 20 pts

ValentinkaMS [17]

Look for the graph where the line crosses the y axis at y=-1, and whose slope is -3. The answer is b.

6 0

3 years ago

The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.

tatuchka [14]

Answer:

a)Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

b) (5211.05 , 5411.7)

95% lower confidence bound on the mean.

c) The test of hypothesis t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Step-by-step explanation:

Step :-1

Given a random sample of 15 devices is selected in the laboratory.

size of the small sample 'n' = 15

An average life of 5311.4 hours and a sample standard deviation of 220.7 hours.

Average of sample mean (x⁻) = 5311.4 hours

sample standard deviation (S) = 220.7 hours.

Step :- 2

a) Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

Level of significance :- α = 0.95 or 0.05

b) The test statistic

$t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }$

$t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }$

t = 5.826

The degrees of freedom γ= n-1 = 15-1 =14

tabulated value t =1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

calculated value t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Null hypothesis is rejected at 95% confidence on the mean.

C) The 95% of confidence limits

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )$

substitute values and simplification , we get

$(5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )$

(5211.05 , 5411.7)

95% lower confidence bound on the mean.

5 0

3 years ago

a tree was 62 inches tall. Three years later, it was 101 inches tall. What was the percent increase, to the nearest hundredth, o

nignag [31]

Answer:

63% increase.

Step-by-step explanation:

You don't really need to do anything because it's probably a 63% increase as 101 is close to 100 already. If you'd like:

$\frac{101-62}{62} *100=$

$\frac{39}{62} *100=$

$62.903225806451612903225806451613$ ≈

63%

8 0

3 years ago