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alukav5142 [94]

3 years ago

7

The length of each side is 3 inches more than the length of each side of a smaller square. The sum of the areas of the squares i

s 425 in^2. Find the lengths of the two sides of the two squares

1 answer:

Margarita [4]3 years ago

4 0

Answer:

Step-by-step explanation:Sides doesn't exist. Given measurements are wrong. Please check your question once again

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For a school fair, Lolita's parents donated 4 bottles of orange juice, 10 bottles of fruit punch and 8 bottles of apple juice.Fi

Vladimir79 [104]

Answer:

2 : 5 : 4.

Step-by-step explanation:

4 : 10 :8

Divide each number by 2:

= 2 : 5 : 4.

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3 years ago

Amy bought a satin ribbon spool containing 131 inches of ribbon. She wants to make eight hair bows from this ribbon. How much ri

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5 0

4 years ago

I need help with this one just give me answer.

PilotLPTM [1.2K]

Answer:

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Step-by-step explanation:

4 0

3 years ago

9/10-?=1/5 please help

zlopas [31]

The answer is 7/10.

5 0

4 years ago

Read 2 more answers

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

degrees of freedom is -1.833 with P = 5%}

P(-1.833 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ ) = 0.95

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ ) = 0.95

<u>95% lower confidence interval for</u> $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $26-1.833 \times {\frac{1.625}{\sqrt{10} } }$ ]

= [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0

3 years ago