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4 years ago

14

On Saturday, mason biked 1.5 hours at a speed of 11.9 miles per hour. on Sunday, he biked 3.2 hours at a speed of 14.8 miles per

hour. how much farther did he bike on Sunday?

1 answer:

solniwko [45]4 years ago

3 0

1.7 miles: subtract 1.5 from 3.2 which equals 1.7 (make sure to line up the decimals!)

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Divide and give the answer in scientific notation:

Nat2105 [25]

It is...
3.0x $10^{-10}$
C

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3 years 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

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

Find all solutions in the interval [0,2pi).<br><br> cos 2x + sqrt(2) sinx=1

kykrilka [37]

<span>cos 2x + sqrt(2) sinx=1
</span><span>
Note that: cos 2x = cos^2x - sin^2x = (1-sin^2x) - sin^2x = 1 - 2sin^2x.
So, when alternatively written, you have the following equation:

</span>- 2sin^2x + sqrt(2)sinx + 1 = 1
- 2sin^2x + sqrt(2)sinx = 0

Then, let z=sin(x). So you get,

- 2z^2 + sqrt(2)z = 0
z(- 2z + sqrt(2)) = 0

Either z=0, or - 2z + sqrt(2) = 0 ---> z=sqrt(2)/2.
Then, since z=0 or z=sqrt(2)/2, therefore sin(x)=0, or sin(x)=sqrt(2)/2.

Then, for you remains just to list the angles. (Let me know if this is not fair or if you got questions.)

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3 years ago

Jodie ran at an average speed of 10 km/h for 15 minutes.<br> How far did she run in km?

laiz [17]

Answer:

2.5 km

Step-by-step explanation:

15 minutes is

15 minutes * 1 hour/ 60 minutes = 1/4 hour

15 minutes = 1/4 of an hour

10 km/ hour * 1/4 hour = 10/4 km = 2.5 km

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3 years ago

Read 2 more answers

Write a division equation to show a 26-pack of candy bars divided equally among 7 people. How many candy bars will each person h

kirza4 [7]

Answer: Each person will get 3 candy bars

Step-by-step explanation:

The equation would be 26/7. Then, 7 goes into 26 three times with a remainder of 5.

$3\frac{5}{7}$

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3 years ago