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3 years ago

Ava is 9 years old. Rose is 11 time as old as ava. How old is rose?

Mathematics

2 answers:

balu736 [363]3 years ago

7 0

99 years old due to multiplying 9 x 11

jek_recluse [69]3 years ago

6 0

99! Since eleven times nine equals 99.

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Someone help with 16!! Please!

Grace [21]

5x-2y=14

2x-y=5

4x-2y=10

x=4

y=3

8 0

3 years ago

4b − 8 = 12. with steps

balandron [24]

4b-8=12
+8 +8

4b=20

Divide both sides by 4

B=5

3 0

3 years ago

Read 2 more answers

Mallanie 2 years older then Bill . Bill is 27 years old

dlinn [17]

really?, 27+2=29, 29 is youre answer.

7 0

4 years ago

Read 2 more answers

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

You are challenged to a lucky draw game. If you draw a face card (K, Q, J) from a standard deck of cards, you earn 10 points. If

sveticcg [70]

There are 3 face cards for each of the 4 suits, leading to a total possible gain of 120, there are 40 other cards leading to a total possible loss of 80.
120 - 80 = 40.
40 / 52 (total number of cards) = about .77

Final Answer:
The expected value of a draw is positive 0.77 points.
Hope I helped :)

7 0

3 years ago

Read 2 more answers