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Katarina [22]
2 years ago
9

1/4x - 5 = 3/4x - 12

Mathematics
1 answer:
Makovka662 [10]2 years ago
3 0

Answer:

Solve for x

x=14

Step-by-step explanation:

by simplifying both sides of the equation, then isolating the variable.

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NEED HELP WITH THIS NEED URGENT PLEASE!!!!!<br> PLEASE ILL MRK BRAINLIEST PLEASE FAST URGENT PLZ!
Schach [20]

Answer:

Table D. shows the proportional relationship.

You would plot those points on their corresponding part on the graph, like (2,1), (6,3), (8,4), and (10,5). The first number being the x value and the second number being the y value. Remember, x goes from side to side, and y goes up and down.

4 0
2 years ago
The following scores were obtained by 11 footballers in a goal-competition 536878311632. The median score was?
sineoko [7]

Answer:

3

Step-by-step:

1,1,2.2.3.3.3.5.6.6.7.8.8

more 3s than other numbers

median is 3

4 0
2 years ago
The small number is a 3 please help
Lunna [17]

Answer:

1

 

Rewrite 200200 as its prime factors.

\sqrt[3]{2\times 2\times 2\times 5\times 5}32×2×2×5×5

2

 

Group the same prime factors into groups of three.

\sqrt[3]{(2\times 2\times 2)\times 5\times 5}3(2×2×2)×5×5

3

 

Rewrite each group of three in exponent form.

\sqrt[3]{{2}^{3}\times 5\times 5}323×5×5

4

 

Use this rule: \sqrt[3]{{x}^{3}}=x3x3=x.

2\sqrt[3]{5\times 5}235×5

5

 

Simplify.

2\sqrt[3]{25}2325

6

 

Rewrite 2525 as {5}^{2}52.

2\sqrt[3]{{5}^{2}}2352

7

 

Use this rule: {({x}^{a})}^{b}={x}^{ab}(xa)b=xab.

2\times {5}^{\frac{2}{3}}2×532

3 0
2 years ago
Read 2 more answers
Dylan wants to invest $3500 for 5 years. The bank offers him 4%/a compounded weekly. Calculate the final amount of this investme
yan [13]

Answer:

the final amount value is $4,274.58

Step-by-step explanation:

Given that

Present value = $3,500

Time period = 5 × 52 = 260

Rate of interest = 4% ÷ 52 = 0.0769%

We need to determine the final amount i.e. future value

So as we know that

Future value = Present value × (1 + rate of interest)^time period

= $3,500 × (1 + 0.0769%)^260

= $4,274.58

Hence, the final amount value is $4,274.58

5 0
3 years ago
A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has
ExtremeBDS [4]

Answer:

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

p_v =P(t_{(23)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=136 represent the sample mean

s=11 represent the sample standard deviation

n=24 sample size  

\mu_o =140 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:  

Null hypothesis:\mu \geq 140  

Alternative hypothesis:\mu < 140  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=24-1=23  

Since is a one left tailed test the p value would be:  

p_v =P(t_{(23)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

4 0
2 years ago
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