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3 years ago

8ftx4ftx5ft what is the diameter

Mathematics

1 answer:

Serggg [28]3 years ago

6 0

Answer:

nothing

Step-by-step explanation:

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Tamara says that 15÷3=5. is she correct? Explain

Alex787 [66]

There are 5 groups of 3 in the number 15, so 15/3=5

4 0

3 years ago

Read 2 more answers

Simplify. Write answers with positive exponents only.

Margaret [11]

Answer:

Step-by-step explanation:

when the base is the same, we subtract when dividing, so

3-2=1, 2x^1=2x

3 0

3 years ago

How do I solve (fraction) 2/3d=1/6?

IgorLugansk [536]

Answer:

d = 1/4

Step-by-step explanation:

Multiply the equation by the inverse of the coefficient of the variable.

$\dfrac{2}{3}d=\dfrac{1}{6}\\\\\dfrac{3}{2}\cdot\dfrac{2}{3}d=\dfrac{3}{2}\cdot\dfrac{1}{6}\\\\d=\dfrac{3\cdot 1}{2\cdot 6}=\dfrac{3}{12}=\dfrac{3\cdot 1}{3\cdot 4}\\\\d=\dfrac{1}{4}$

8 0

3 years ago

Please help it’s due now

Elan Coil [88]

Answer:

increase

Step-by-step explanation:

increase because it's going up

8 0

3 years ago

Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of <em>t</em> for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

6 0

3 years ago