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Annette [7]
2 years ago
9

(5x - 18) + (4x + 7) = 180 what is the value of x?

Mathematics
1 answer:
CaHeK987 [17]2 years ago
7 0
(5x-18)+(4x+7)=180
5x+4x=180+18-7
9x=191
x=191/9

Answer: x=191/9
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Which are the solutions of x2 = –11x + 4? StartFraction negative 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFr
Marysya12 [62]

Answer:

x_1=-\frac{11}{2}-\frac{\sqrt{137} }{2}\\\\x_2=-\frac{11}{2}+\frac{\sqrt{137} }{2}

Step-by-step explanation:

Given the following quadratic equation:

x^2 = -11x + 4

The steps to solve it are:

1. Move the terms to one side of the equation:

x^2+11x- 4=0

2. Apply the Quadratic formula x=\frac{-b\±\sqrt{b^2-4ac} }{2a}.

In this case we can identify that:

a=1\\b=11\\c=-4

Then, substituting these values into the Quadratic formula we get the following solutions:

x=\frac{-11\±\sqrt{11^2-4(1)(-4)} }{2(1)}

x_1=-\frac{11}{2}-\frac{\sqrt{137} }{2}\\\\x_2=-\frac{11}{2}+\frac{\sqrt{137} }{2}

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The operations manager of a manufacturer of television remote controls wants to determine which batteries last the longest in hi
Margarita [4]

Answer:

t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

The degrees of freedom are given by:

df=n_{1}+n_{2}-2=4+4-2=6

The p value for this case would be given by:

p_v =2*P(t_{(6)}

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

Step-by-step explanation:

Information given

Battery 1 106 111 109 105

Battery 2 125 103 121 118

We can calculate the mean and the deviation with the following formulas"

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_{1}=107.75 represent the mean for the Battery 1

\bar X_{2}=116.75 represent the mean for the Bettery 2

s_{1}=2.75 represent the sample standard deviation for the Battery 1

s_{2}=9.604 represent the sample standard deviation for the battery 2

n_{1}=4 sample size selected for the Battery 1

n_{2}=4 sample size selected for the Battery 2

\alpha=0.1 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

Null hypothesis:\mu_{1} = \mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

The statistic is given by:

t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

The statistic is given by:

t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

The degrees of freedom are given by:

df=n_{1}+n_{2}-2=4+4-2=6

The p value for this case would be given by:

p_v =2*P(t_{(6)}

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

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Answer:

Step-by-step explanation:

20

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Least to greatest - 1.8 , 1 , 1 2/5 , 1 9/10 , -1.25
nika2105 [10]

Answer:

-1.25, 1, 1 2/5, 1.8, 1 9/10

Step-by-step explanation:

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