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STALIN [3.7K]
3 years ago
15

On a string of lights there are 200 bulbs. After many years of use, 19 of them have burned out. Approximately what percentage of

bulbs have burned out?
Mathematics
1 answer:
Allisa [31]3 years ago
8 0

Answer:

8.5% (you can approximate that, that's the exact number and it isn't too funky)

Step-by-step explanation:

Percentage is out of 100, and over here we see that 200 is 2 times the amount of 100. So that means that if we divide 200 by 2, we get 100, which means we also divide 19/2 to get 8.5. That means, out of 100 light bulbs, 8.5 of them burned out. This is really easy to convert because the percentage is already out of 100, so you just make 8.5 into a percent which is 8.5%

Let me know if you have any questions.

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Need asap final exam help fast
antiseptic1488 [7]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

7 0
3 years ago
Read 2 more answers
2. What is the value of x?<br> A. 17<br> B. 39<br> C. 26<br> D. 41
creativ13 [48]
41 a squared plus b squared equals c squared so it would be 41 have a nice day
8 0
3 years ago
An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample
valentina_108 [34]

Answer:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Replacing we got:

\bar X=58.875 represent the mean

s=5.083 represent the sample standard deviation for the sample  

n=8 sample size  

\mu_o =60 represent the value that we want to test

\alpha=0.1 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:  

Null hypothesis:\mu \geq 60  

Alternative hypothesis:\mu < 60  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info we got:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0
3 years ago
With the drain open, a pool loses water at a rate of 9 gallons per minute. At that rate, how long will it take to drain 486 gall
Snezhnost [94]

divide 486 by 9 = 54

it would take about 54 minutes

8 0
3 years ago
I'm not sure what's the answer
kow [346]

Answer:

m=17/40  \frac{17}{40}

Step-by-step explanation:

So you 5/7m - 1/7 = 9/56 \frac{5}{7}m - \frac{1}{7} = \frac{9}{56}

You Simplify both sides Then

well I Don´t know I hope this helped u, I tried ^^¨

8 0
4 years ago
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