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omeli [17]
3 years ago
5

twice the difference between a number and 10 is equal to 6 times the number plus 14. what is the number?

Mathematics
1 answer:
laiz [17]3 years ago
4 0
Let the number be n.  Then 2(n-10) = 6n + 14.

Multiplying out the left side, 2n - 20 = 6n + 14.  Thus, -34=4n, and n = -17/2.
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A. 1/5

Step-by-step explanation:

Add all the numbers up, you get 25. There is 5 yellow marbles, so that's a 5/25 chance of picking a yellow one, now you just have to simplify, you then get 1/5, because 5 ÷ 5 is 1, and 25 ÷ 5 is 5.

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Step-by-step explanation:

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On her birthday, Sonia distributed chocolates in an orphanage. She gave 5 chocolates to each child and 20 chocolates to adults.
qwelly [4]

Answer with explanation:

Total number of Children = x Children

Total number of chocolates distributed = y Chocolates

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Let total number of adults be z.

Expressing the statement in terms of Equation

 y = 5 x + 20 z

z=\frac{y-5 x}{20}

⇒Number of Adults =z

       =\frac{y-5 x}{20}

3 0
3 years ago
PLEASE HELP ME 20 POINTS + BRAINLIEST
nataly862011 [7]

9514 1404 393

Answer:

  (b)  36.56 ft

Step-by-step explanation:

The perimeter is the total of the circumference of a 4 ft circle and two lengths that are 12 ft.

The circumference is given by ...

  C = πd

  C = (3.14)(4 ft) = 12.56 ft

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8 0
3 years ago
Read 2 more answers
Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp
aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}

Using a statistic software I've calculated the test

F_{H_0}= 1.41

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0
3 years ago
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