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3 years ago

Which of the following is the best description of a positive association between two variables?

Mathematics

1 answer:

julsineya [31]3 years ago

4 0

B) the values will create a line with positive slope when graphed on a scatter plot.

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A prop for the theater club’s play is constructed as a cone topped with a half-sphere. What is the volume of the prop? Round you

marta [7]

Volume of cone formula: V = PI x r^2 x h/3

Volume of sphere formula: 4/3 x PI x r^3

Cone: 3.14 x 9^2 x 9/3 = 763.02 cubic inches

1/2 sphere: 1/2 x 4/3 x 3.14 x 9^3 = 1526.04 cubic inches.

Total volume : 1526.04 + 763.02 = 2289.1 cubic inches

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3 years ago

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Liono4ka [1.6K]

Answer:

Step-by-step explanation:

Area of H = 25pi

Area of J = 225pi

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3 years ago

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What is the y intercept

Gemiola [76]

Answer:

Y intercept is the value of y when x is equaled 0.

Step-by-step explanation:

Slope intercept form is

y = mx + b

In this equation, b is the y intercept.

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2 years ago

Which of the following statements about the box-and-whisker plot below is not true?

Alex_Xolod [135]

Answer: D

Step-by-step explanation:

Because the Midrange data set shown is 55

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3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

3 years ago