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Semmy [17]
3 years ago
6

44​% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorit

e nut. Find the probability that the number who say cashews are their favorite nut is​ (a) exactly​ three, (b) at least​ four, and​ (c) at most two. If​ convenient, use technology to find the probabilities.
Mathematics
1 answer:
blondinia [14]3 years ago
6 0
The answer to this is at least 4
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What is -5/4×1/3 and how do you solve fractions??​
kolbaska11 [484]
First off when multiplying fractions you always multiply straight across. so the answer would be -5/12.
6 0
3 years ago
A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time).
likoan [24]

Answer:

(a). 72.9%.

(b). 13.6 hr.

Step-by-step explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this question/problem;

=> "A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time). "

=> The worker produces 10 welded assemblies on the second day, and the 10th assembly on the second day takes 30 minutes"

So, we will be making use of the Crawford learning curve model.

T(7) + 10 = T (17) = 30 min.

T(7) = T1(7)^b = 45.

T(17 ) = T1(17)^b = 30.

(T1) = 45/7^b = 30/17^b.

45/30 = 7^b/17^b = (7/17)^b.

1.5 = (0.41177)^b.

ln 1.5 = b ln 0.41177.

0.40547 = -0.8873 b.

b = - 0.45696.

=> 2^ -0.45696 = 0.7285.

= 72.9%.

(b). T1= 45/7^ - 045696 = 109.5 hr.

V(TT)(17) = 109.5 {(17.51^ - 0.45696 – 0.51^ - 0.45696) / (1 - 0.45696)} .

V(TT) (17) = 109.5 {(4.7317 - 0.6863) / 0.54304} .

= 815.7 min .

= 13.595 hr.

8 0
3 years ago
HELP URGENT!! ILL GIVE BRAINLIEST
Luba_88 [7]

Answer:

I think its 21.34 sorry if i am wrong

6 0
2 years ago
Read 2 more answers
A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class
KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

<u><em /></u>

<u><em>Let p = percentage of seniors who skip their classes.</em></u>

So, Null Hypothesis, H_0 : p \leq 2%   {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, H_A : p > 2%   {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                   T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of seniors who skipped their classes = \frac{77}{2521}

           n = sample of classes = 2521

So, <u><em>test statistics</em></u>  =  \frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }

                               =  3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, <u>the z table gives critical value of 2.3263 for right-tailed test</u>. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that seniors skip more than 2% of their classes.

6 0
3 years ago
HELP ASAP
balandron [24]

56

Step-by-step explanation:

The whole angle is 132 and half of it is 76, so the other portion would be 56 i believe.

4 0
2 years ago
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