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dimulka [17.4K]
3 years ago
12

A researcher finds a significant correlation between the amount of time spent watching television and blood pressure for a sampl

e of 40-year-old men. This means that watching more television causes high blood pressure.
O True

O False
Mathematics
1 answer:
kobusy [5.1K]3 years ago
8 0

Answer:

statement is false

Step-by-step explanation:

given data

men age = 4 year old

relation between = watching television and blood pressure

solution

we can say that this statement is wrong

because here causality it have never implied by correlation between these two watching television time period and the blood pressure as that by watching more time television that not causes high blood pressure

so that statement is false

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Drupady [299]

。☆✼★ ━━━━━━━━━━━━━━  ☾

Divide by 2 to get 1/3 of the week:

1/4 / 2 = 1/8

Multiply by 3 to get the full week:

1/8 x 3 = 3/8

Thus your answer is B. 3/8

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8 0
4 years ago
Pythagorean theorem in 3D.
White raven [17]

Answer:

<em>h = 6.2</em>

Step-by-step explanation:

You need to find h.

The dashed triangle is a right triangle. 8 is the hypotenuse. h is a leg. The other leg is the half of the side of the base, 5.

5^2 + h^2 = 8^2

h^2 + 25 = 64

h^2 = 64 - 25

h^2 = 39

h = \sqrt{39}

h = 6.244998...

h = 6.2

6 0
3 years ago
When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted
hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

<em>H₀</em>: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

Assume that the significance level of the test is, <em>α</em> = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}

Compute the pooled standard deviation as follows:

S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16

The critical value of <em>t</em> is:

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984

*Use a <em>t</em>-table.

Compute the 95% confidence interval for mean difference as follows:

CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}

     =(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}

     =-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0
4 years ago
Find the measure of YVZ
Arisa [49]

Answer:

x 100

Step-by-step explanation:

because of the yvz angle its equals 100-12

5 0
4 years ago
A JIT system uses kanban cards to authorize movement of incoming parts. In one portion of the system, a work center uses an aver
mrs_skeptik [129]

Answer: 6 Containers are needed

Step-by-step explanation:

The ideal number of Kanban cards can be calculated using the following formula;

N = (DT × ( 1 + X)) / C

where N is total number of container,

D is panned usage rate of using work Centre

T is Average waiting time for replenishment of parts plus average production for a  container of parts,

X is policy variable set by management that reflects possible inefficiency in the system ,

C is capacity of standard container.

Now, given that;

D = 105 parts

T = 90min = 90/60 = 1.5 hrs

X = 0.35

C = 3 × 12parts = 36 parts

so; we substitute

N = (105×1.5 × ( 1 + 0.35)) / 36

N = 212.626 / 36

N = 5.9063 ≈ 6

Therefore 6 Containers are needed

8 0
3 years ago
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