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ss7ja [257]
3 years ago
6

The perimeter of a rectangular field is 298 yards. If the width of the field is 63 yards, what is its length?

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

The length is 86 yards.

Step-by-step explanation:

63 is the width so,

63 + 63 = 126

Now subtract this from the total perimeter,

298 - 126 = 172

and divide by 2 because there are sides with the same length

172/2 = 86

The length is 86.

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2x – 3 &gt; -4x + 2<br> What is the solution to the inequality
bagirrra123 [75]

Answer:

x>\frac{5}{6}

Step-by-step explanation:

2x-3>-4x+2

Add 3 to both sides

2x-3+3>-4x+2+3

Simplify

2x>-4x+5

Add 4x to both sides

2x+4x>-4x+5+4x

Simplify

6x>5

Divide both sides by 6

\frac{6x}{6} >\frac{5}{6}

Simplify

x>\frac{5}{6}

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D. (3+5i) (6-i) = Show your work.​
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Answer:

4i + 9 < simplified version

Step-by-step explanation:

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PLEASE HELP ASAP!!!!! WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!
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The equation of the graph show is 

y = -2/3x - 2
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2x4^7=<br>What is 2x4^7<br>Example: 2x5^7=2x=78,125
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A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of
finlep [7]

Answer:

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

p_v =2*P(t_{(7)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=0.475

The sample deviation calculated s=0.183

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 0.5  

Alternative hypothesis:\mu \neq 0.5  

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{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

P-value

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

df=n-1=8-1=7  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

4 0
3 years ago
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