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3 years ago

-(7-4x)=9can somebody help me with this, and if you could, can you show work?

Mathematics

2 answers:

Alex777 [14]3 years ago

4 0

Answer:

x=4

Step-by-step explanation:

Simplifying

-1(7 + -4x) = 9

(7 * -1 + -4x * -1) = 9

(-7 + 4x) = 9

Solving

-7 + 4x = 9

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '7' to each side of the equation.

-7 + 7 + 4x = 9 + 7

Combine like terms: -7 + 7 = 0

0 + 4x = 9 + 7

4x = 9 + 7

Combine like terms: 9 + 7 = 16

4x = 16

Divide each side by '4'.

x = 4

luda_lava [24]3 years ago

3 0

Answer:

x = 4

Step-by-step explanation:

-(7 - 4x) = 9 (Given)

-7 + 4x = 9 (Distributed)

-7 + 4x + 7 = 9 + 7 (Add 7 on both sides)

4x = 16 (Simplify)

4x/4 = 16/4 (Divide 4 on both sides)

x = 4 (Simplify)

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Point A is located at (5, 10) and point B is located at (20, 25).

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Let's say the point is C, so C partitions AB into two pieces, where AC is at a ratio of 3 and CB is at a ratio of 7, thus 3:7,

$\bf \left. \qquad \right.\textit{internal division of a line segment}
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A(5,10)\qquad B(20,25)\qquad
\qquad 3:7
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\cfrac{AC}{CB} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(5,10)=3(20,25)\\\\
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{ C=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\
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3 years ago

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

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Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

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 deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

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3 years ago

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