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2 years ago

Determine if they are congruent. State how they are.

Mathematics

1 answer:

Doss [256]2 years ago

5 0

$\huge \bf༆ Answer ༄$

The Correct choice is ~ D

$\textsf{Not enough information }$

The given triangles are not congruent, because they they just have all their corresponding angles equal to one another, so we need more information ~

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How to solve 38.7×10 to the fourth power

Sergeu [11.5K]

Answer: 3.87 * 10^5, or 387000

Step-by-step explanation:

Move the decimal to the left and increase the power.

If you want to have it expanded, move the decimal point to the right 4 times of the original to get 387000.

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4 years ago

How many tens are in 100

soldier1979 [14.2K]

There are 10 tens in 100

4 0

3 years ago

Read 2 more answers

Write 0.002358 in scientific notation.

statuscvo [17]

The answer is
2.358×10^-3

6 0

4 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

BRAINLIEST + POINTS<br><br> Can someone explain why is B correct? Please show your work

Viefleur [7K]

Answer:

B. $1680

Step-by-step explanation:

Note the simple interest formula:

Interest = (Principal)(rate)(time)

Principal = $6000

Rate = 7% = 0.07

Time = 4

Plug in the corresponding numbers for the corresponding variables.

I = (P)(R)(T)

I = (6000)(0.07)(4)

Multiply.

I = (24000)(0.07)

I = 1680

Interest is $1680.

5 0

4 years ago