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2 years ago

Ella’s geometry teacher asked each student to devise a problem and write out its solution. Here is Ella’s work:

Mathematics

2 answers:

Vaselesa [24]2 years ago

7 0

Answer:

Ellas procedure and conclusion are incorrect

Step-by-step explanation:

Marta_Voda [28]2 years ago

4 0

Answer:Ella's Procedure And Conclusion are incorrect

Step-by-step explanation:

You might be interested in

You offer senior citizens a 20% discount on their tune-ups at your gas station. Assuming that an average of 50 senior citizens g

andreyandreev [35.5K]

Percentage of discount given to senior citizens for tune-ups at the gas station = 20%
Average number of senior citizens visiting the gas station per month = 50
Average price of tune-up before discount = $49.95
Total amount that would
have been collected before discount = 49.95 * 50
= 2497.50 dollars
Total amount of discount given = (20/100) * 2497.50
= 499.50 dollars
I hope that the procedure is clear enough for you to understand.

6 0

2 years ago

I would like to purchase 20 products at a cost 65.00 per product. what will be my total? your state has 3.5% sales tax

xxTIMURxx [149]

20*65 = 1300
That's the price without tax.
Now to find tax:
1300*.035 = 45.5
1300+45.5 = $1345.5

$1345.5 is your total

5 0

2 years ago

Read 2 more answers

The function f(x) = –10(x)(x – 4) represents the approximate height of a projectile launched from the ground into the air with a

sineoko [7]

The height of the project becomes zero when it hits the ground which is also the total time at which the projectile stays in the air. Equating the function to zero:
-10(x)(x - 4) = 0
The roots of this quadratic equation are:
x = 0, and x=4

Therefore, the projectile stays in the air for
4 seconds

4 0

2 years ago

Read 2 more answers

A train is traveling at a constant speed and has traveled 67.5 miles in the last 11 hours.

Rudiy27

Answer:

A. d= 45t

Step-by-step explanation:

(assuming that you meant 67.5 in the last 1.5 hours)

67.5 miles = distance

1.5 hours = time

therefore:

$\frac{d}{t\\}$ = 67.5/1.5

making your answer 45

leaving a as your correct answer:

d= 45t

3 0

1 year ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries

Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = $\frac{\sum X}{n}$ = 26 hr

s = sample standard deviation = $\sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} }$ = 1.625 hr

n = sample of batteries = 10

$\mu$ = population mean battery life

<em> Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, $\mu$ is;

P(-1.833 < $t_9$ ) = 0.95 {As the lower critical value of t at 9