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babymother [125]
3 years ago
12

Please help I have school

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

Answer:

77

Step-by-step explanation:

8 and 2/5 times 18 and 1/3 is 154. Divide 154 by 2 or multiply by 1/2 and get 77

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Which is the correct solution to the expression 4^8? Use the scientific calculator given in this course.
Firdavs [7]
4^8 = 4\times 4\times 4\times 4\times 4\times 4\times 4\times 4 = \boxed{\bf{65536}}

Your final answer is 65,536.
5 0
4 years ago
Read 2 more answers
A teacher has a total of 7/9 bucket of water to pour into pots with flowers that are standing in equal quantities on two windows
AleksAgata [21]

Answer:

14 pots

Step-by-step explanation:

if teacher pours 1/18 in each pot and there was 7/9 buckets in total, 9 fits into 18 twice so just 7X2 =14 and 9X2=18 so 14 pots were watered

hope this helps

3 0
3 years ago
Chris tries to copy
serious [3.7K]

Answer:

Man what ????

Step-by-step explanation:

MAN WHAT ????

4 0
4 years ago
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Find x , if someone can help me with all my hw please lmk
choli [55]

Answer:

ok so if we look at this other one we can see that the other angle is 45

since we now all three angles we can find the other side

Side a = 4

Side x = 5.65685 = 4√2

Side c = 4

5.66

Hope This Helps!!!

4 0
3 years ago
Read 2 more answers
Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The m
svp [43]

Answer:

As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.

Step-by-step explanation:

As we suspect the variability of concentration F - test is applied.

n1=10    s1=4.7

n2=16      s2=5.8

And α = 0.05.

The null and alternate hypothesis are

H0: σ₁²=σ₂²       Ha:  σ₁²≠σ₂²

The null hypothesis is  the variability in concentration does not  differ for the two companies.

against the claim

the variability in concentration may differ for the two companies

The critical region F∝(υ1,υ2) = F(0.025)9,15= 3.12

and 1/F∝(υ1,υ2) = 1/3.77= 0.26533

where υ1= n1-1= 10-1= 9 and υ2= n2-1= 16-1= 15

Test Statistic

F = s₁²/s₂²

F= 4.7²/5.8²=0.6566

Conclusion :

As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.

3 0
3 years ago
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