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weqwewe [10]
3 years ago
9

It is known that 11% of the seniors in a large high school enter military service upon graduation. If a group of 10 seniors are

randomly selected, what is the probability of observing exactly two who will be entering military service?
Mathematics
1 answer:
GaryK [48]3 years ago
6 0

Answer:

2.2%

Step-by-step explanation:

I don't know if this right but it would be 2/10 and multiply htat by 0.11 (representing the 11%) and you would get 0.022 or 2.2%

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If the weight of a class 4 truck is increased by two tons will it still be classified as a class 4 truck
Assoli18 [71]
If the truck is 14,000 pounds and increases by 2 tons yes.  However if it is above 14,000 then no it will not be a class 4 truck.
8 0
3 years ago
A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o
Mandarinka [93]

Answer:

(a) Null Hypothesis, H_0 : \mu = 22 ounces  

    Alternate Hypothesis, H_A : \mu\neq 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject H_0 since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

<em>Let </em>\mu<em> = average ounces of passion fruit juice used to fill each bottle.</em>

(a) So, Null Hypothesis, H_0 : \mu = 22 ounces     {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, H_A : \mu\neq 22 ounces     {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                        T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean weight of the passion fruit juice = 21.54 ounces

            \sigma = population standard deviation = 1.38 ounce

            n = sample of bottles = 65

So, <u><em>test statistics</em></u>  =  \frac{21.54-22}{\frac{1.38}{\sqrt{65} } }  

                               =  -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

<em>Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that Reject H_0 since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

6 0
3 years ago
Shirley cuts a 133 m long rope into several pieces of length 3 1/2 m each. How many equal pieces of the rope does she get
liraira [26]


38

explanation:
133 divided by 3.5
5 0
3 years ago
kelly makes $10 more an hour than her regular pay when she is working overtime. Yesterday, she worked 8 regular hours and 4 over
skad [1K]
C. $10 per regular hour.
5 0
3 years ago
PLEASE HELP! ANY SPAM ANSWERS WILL BE REPORTED!
Rainbow [258]

Answer:

Rise: -1   Run: -1

Step-by-step explanation:

It's a negative line, and it crosses at the corner of every line. Hope this helps :)

7 0
2 years ago
Read 2 more answers
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