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baherus [9]
3 years ago
14

Which is smaller, one-ninth of 54 or one-fourth of 36?

Mathematics
2 answers:
Julli [10]3 years ago
8 0

Answer:

1/9 of 54  is smaller than 1/4 of 36

Step-by-step explanation:

1/9 of 54 is 54/9 and that is 6

1/4 of 36 is 36/4 and that is 9.

Please mark brainliest

RideAnS [48]3 years ago
6 0

Answer:

one-ninth of 54

Step-by-step explanation

54*1/9=6

36*1/4=9

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The reciprocal of 9 2/3 is 3/29

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Solve the equation : x^(2)+5=3(x^(2)-3)
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x = ± 2

Step-by-step explanation:

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x² + 5 = 3x² - 3 ( subtract x² + 5 from both sides )

0 = 2x² - 8 ( add 8 to both sides )

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What’s the LCM for 22, 44 show your work
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What is the value of 0.6 + 0.3? Write your answer as a fraction in lowest terms
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3 0
3 years ago
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
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