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musickatia [10]
3 years ago
6

What is the quotient of a number and three is negative four

Mathematics
1 answer:
Alecsey [184]3 years ago
3 0

Answer:

-7

Step-by-step explanation:

7-3=4

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A sample of 14001400 computer chips revealed that 311% of the chips do not fail in the first 10001000 hours of their use. The co
Archy [21]

Answer:

There is sufficient evidence at 0.05 significant level to support company's claim.

Step-by-step explanation:

We have these informations from the question

n = 1400

P^ = 31% = 0.31

Alpha level = 5% = 0.05

Then we come up with the hypothesis

H0: P = 0.28

H1: P>0.28

From here we calculate the test statistic

z = p^ - p/√pq/n

P = 0.28

q = 1-0.28

= 0.72

z = 0.31-0.28/√(0.31*0.72)/1400

= 0.03/√0.0001594

= 0.03/0.012

= 2.5

Then we have a p value = 0.00621

The p value is less than significance level

0.00621<0.05

So the null hypothesis is rejected.

We conclude that There is sufficient evidence at 0.05 significant level to support company's claim.

5 0
2 years ago
A right triangular prism and its net are shown below.
Law Incorporation [45]

Answer:

A=2

B=12

C=5

D=13

SA=411mm

Step-by-step explanation:

SA=bh+(s1+s2+s3)H

=12×5+(12+2+13)(13)

6 0
2 years ago
Read 2 more answers
What is the pre-image vertex A’ if the rule that created the image is r y-axis (X,Y) (-x,y)
Lapatulllka [165]

Answer:

its A=(-4,2)

Step-by-step explanation:

4 0
3 years ago
What is 25 more than n in an algebraic expression
sashaice [31]
25 + n
////////////////
8 0
3 years ago
An experiment was conducted to compare the wearing qualities of three types of paint (A, B, and C) when subjected to the abrasiv
MaRussiya [10]

Answer:

A) H0:  mean of A= mean of B= mean of C

   H1: all means are unequal

B) 3.48194

C) 0.04515

D) H0 is accepted. There is significant difference among mean hours after which visibile abrasion is observed for each type of paint

Step-by-step explanation:

A) Here Anova test will be used as we are comparing three groups

B) mean of A= 229.6, SD of A= 158.1962, SE of A= 50.026

   mean of B= 309.9 , SD of B= 147.8742, SE of B= 46.7619

   mean of C= 427.8, SD of C= 196.8179, SE of C= 62.2393

overall mean= 322.43

degrees of freedom between groups= 3-1=2

degrees of freeddom of error= total number of observations - number of groups= 30-3= 27

total degrees of freedom= total number of samples-1= 30-1 =29

Sum of squares between groups=∑ sample size of group×(individual group mean- over all mean)²

=198772.4667

mean of sum of squares between groups= sum of squares between groups/ (number of groups -1)

  = 198772.4667/2

sum of squares of error=∑∑ (individual observation of group- group mean)²

                                            = 770670.9223

mean sum of squares of erros= sum of squares of error/(total number of samples-number of groups)

                                                       = 770670.92/27

                                                         = 28543.38

F- stat= mean sum of squares between groups/ mean sum of squares of error

     F stat= 3.48194

C) p- value= 0.04515

D) since F-stat is greater than p- value, null hypothesis is rejected

from F distribution table

F(2,27)=

3 0
3 years ago
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