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lozanna [386]
2 years ago
14

Researchers found that a person in a particular country spent an average of 5.5 hours a day watching TV in 2010. Assume the popu

lation standard deviation is 1.8 hours per day. A sample of 39 people averaged 6.3 hours of television viewing per day. Does this result support the findings of the​ study?
Mathematics
1 answer:
zmey [24]2 years ago
5 0

Answer:

Yes

Step-by-step explanation:

Because 6.3hours is still in the range 5.5+/-1.8 hours

Therefore the result support the findings of the study.

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10) A cube has width of 10 cm. What is the volume of the cube?
PSYCHO15rus [73]

Volume of the cube is 1000 cm³

Step-by-step explanation:

  • Step 1: Find the volume of the cube.

Side of the cube = 10 cm

⇒ Volume of the cube  =(side)³ = 10³ = 1000 cm³

7 0
3 years ago
Read 2 more answers
A study was conducted to determine whether an expectant mother's cigarette smoking has any effect on the bone mineral content of
svp [43]

Using Hypothesis testing,

a) Two samples are independent.

b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.

Hₐ : an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

c) Null hypothesis is accepted.

so, an expectant mother's cigarette smoking has no effect on the bone mineral.

We have given that,

A study which is conducted for check an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

For new-born whose mother's cigarette smoking

sample size , n₁= 77

X-bar, x₁-bar = 0.098 g/cm

standard deviations, s₁ = 0.026 g/cm

For new-born whose mother did not cigarette smoking

sample size , n₂ = 161

standard deviations, s₂ = 0.025 g/cm.

mean (X-bar) , x₂-bar = 0.095 g/cm

a) the two samples are independent since they are different types of mothers, smoking mothers and non smoking mothers

b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.

Hₐ: an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

c) Test statistic:

t = (x₁-bar - x₂-bar )/S (√1/n₁+1/n₂)

where , S = √(s₁²(n₁ - 1) + s₂²(n₂ -1))/n₁+n₂ - 2

S = √((0.026)²(76) +(0.025 )²(160))/236

= 0.0253

then, t = (0.098 - 0.095 )/0.0253(√1/77 +1/161 )

=> t = 0.859

Using the critacal table critical t is

Critical t = ±1.970065

Degrees of freedom =236.0000

P-Value=0.3935 which is greater than α(0.05),

So , we accept H₀

Thus, an expectant mother's cigarette smoking has not effect on the bone mineral content of her otherwise healthy child.

To learn more about Hypothesis testing, refer:

brainly.com/question/4232174

#SPJ4

8 0
1 year ago
BRAINLIEST ASAP! PLEASE HELP ME :)
Afina-wow [57]

Answer:

\large \boxed{\textbf{30.54 yr}}

Step-by-step explanation:

For species A,

P =2000e^{0.05t}

For species B,

P =5000e^{0.02t}

When their populations are equal,

\begin{array}{rcl}2000e^{0.05t} & = & 5000e^{0.02t}\\e^{0.05t} & = & 2.5e^{0.02t}\\0.05t & = & \ln2.5 + 0.02t\\0.05t & = & 0.9163 + 0.02t\\0.03t & = & 0.9163\\t& = & \dfrac{0.9163}{0.03}\\\\& = & \mathbf{30.54}\\\end{array}\\\text{The populations of species A and B will be the same in $\large \boxed{\textbf{30.54 yr}}$}

The diagram below shows the population of Species A overtaking that of Species B in about 30.5 yr.

3 0
3 years ago
As a​ follow-up to a report on gas​ consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage f
kondor19780726 [428]

Answer:

The 95​% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).

Step-by-step explanation:

Our sample size is 96.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 96-1 = 95

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 95 and 0.025 in the t-distribution table, we have T = 1.9855.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{5.6}{\sqrt{96}} = 0.57

Now, we multiply T and s

M = T*s = 1.9855*0.57 = 1.31

For the lower end of the interval, we subtract the mean by M. So 19.6 - 1.31 = 18.29

For the upper end of the interval, we add the mean to M. So 19.6 + 1.31 = 20.91

The 95​% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).

8 0
3 years ago
There are20 keys on a keychain. One of the keys starts a car. If a key is randomly chosen what is the probability that it starts
pogonyaev
The answer is 5%........


7 0
3 years ago
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