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4 years ago

10

Two integers are some at random integers one through 15. Find the probability that both the numbers are even if they are not rep

laced

1 answer:

sasho [114]4 years ago

3 0

Hope this helps let me know if does

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The circumference of a circle is 14 cm. Find the radius of this circle

Dafna1 [17]

2.23 should be the radius

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3 years ago

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Answer to three friends each create 4 bags of starter bread dough. After ten days, each of those four bags is then divided into

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<span>To begin, each of the friends have made 4 bags of dough. This gives 12 bags at the outset. After 10 days, all 12 bags are made into (4*12) bags, or 48 bags. After 20 days, the 48 bags are made into (48*4), or the required 192 bags. After 20 days, then, the friends have the needed 192 bags.</span>

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3 years ago

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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:

a

C

b

C

c

C

d

A

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

Please help me with this on the image

alukav5142 [94]

Answer:

200 grams

Step-by-step explanation:

80 grams x grams

-------------- = -------------

6 people 15 people

Using cross products

80 *15 = 6x

1200 = 6x

Divide by 6

1200/6 = 6x/6

200 =x

8 0

3 years ago

What is five and three fourths plus one and seven tenths. Answer choices: seven and nine twentieths, seven and one twentieth, si

MrMuchimi

Answer:

5 3/4+1 7/10

23/4+17/10

<u>115</u><u>+</u><u>34</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u>20</u>

<u>=</u>149/20

=7 9/20

Step-by-step explanation:

the answer is seven and nine twenntieths

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3 years ago

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