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-Dominant- [34]
3 years ago
10

What is the LCM and GCF of 37 and 12

Mathematics
1 answer:
RSB [31]3 years ago
7 0
I hope this helps you

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Please help I need this immediately
svetlana [45]

Answer:

The choose B. x = 7 , x = – 15

(x+4)-5=6

x +4-5=6 —> x=6-4+5 —> x= 7

– (x+4)-5=6 —> -x -4-5=6—> x = -4 -5 -6 —> x = – 15

I hope I helped you^_^

8 0
3 years ago
3. Kirk bought a bag of candy and took 10
tamaranim1 [39]
70, Multiply 5 pieces by 12 friends and you get 60, then add the 10 pieces he originally took out, and you get 70
8 0
3 years ago
HELP PLSSSSSSSSSSSSSSS ASAP
makvit [3.9K]

Answer:

26

Step-by-step explanation:

Sydney = x

Devaughn = 2x

x + 2x = 78

3x = 78

x = 78 / 3

x = 26

Hope that helps!

6 0
2 years ago
Read 2 more answers
Media experts claim that daily print newspapers are declining because of Internet access. Listed​ below, from left to right and
yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic  t_{\alpha/2} is 6.678155

(d) The p value is  1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median  

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean,  \bar{x}_{1}= 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median  

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean,  \bar{x}_{2}}= 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic  t_{\alpha/2} = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

5 0
3 years ago
HELP PLZ<br><br> 7k + 4s = 20<br> -8k + 2s = 10<br><br> What is the value of k?
Finger [1]
Divide 20 by 4=5
then 5 by 7=5/7

k=5/7
8 0
3 years ago
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