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liq [111]
3 years ago
7

Can someone help me with this question.

Mathematics
2 answers:
ioda3 years ago
6 0

Answer:

The answer to your problem is Option A or 1: A football team gains 15 yards on a play.

Step-by-step explanation:

This is the answer because all of the other situations have an integer of -15. Like LOSING points, or 15 meters BELOW all signify that those situations include a negative number. Your question is asking for a positive 15 not negative so Option 1 is the correct answer.

NISA [10]3 years ago
3 0

Answer:

Option A

Step-by-step explanation:

If a football team gains 15 yards on a play, it will be recorded as '+ 15' or '15'.

All other options would be represented by '-15', and not '15' that is asked in the question. A withdraw is taking away money, a number 'below zero' would be negative, and losing 15 points would be represented as '-15'.

Option A should be the correct answer.

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2 over 12 girls  in all.
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Determine sen 0 y tan 0,si 0= 2/3 y cot 0>0
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the answer of your question is 0

Step-by-step explanation:

the answer is 0

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The mass of Hinto's math is 4,367 grams. What is the mass of 5 math books in kilograms?
suter [353]

Answer:

21.835 kilograms

Step-by-step explanation:

(4,367x5)/1000

7 0
3 years ago
At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

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=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

8 0
3 years ago
The sum of ten consecutive whole even numbers is 1190
olga2289 [7]
X + x+2 + x+4 + x+6 +x+8 + x+10 + x+12 + x+14 + x+16 + x+18 = 1190
Now simplify
10x + 90 = 1190
10x = 1100
x = 110
Now plug in 110 for x
5 0
3 years ago
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