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weeeeeb [17]
3 years ago
12

Suppose you have developed a scale that indicates the brightness of sunlight . Each category in the tablet is 5 time brighter th

an the category above it. For example, a day that is dazzling is 5 times brighter than a day that is radiant . How many times brighter is a dazzling day than a dim day
Dim is 2
Mathematics
2 answers:
Ostrovityanka [42]3 years ago
8 0

If you could put in the tablet.

But the equation would be: a_n = a_1 (r)^(x-1)

with a_1 being the first category

r being 5 how much brighter than the day before is

and x is what nr. in the sequence it is

But I'm not sure as I cannot see the table

umka21 [38]3 years ago
4 0

Answer:

10 Times

Step-by-step explanation:

From the statement "Each category in the tablet is 5 time brighter than the category above it." It can be inferred that the brightest portion of the day must be 5 times the less bright day. This means that:

Dim = 2

Brightest = 5 times dim

               = 5 * 2

               = 10

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A laptop producing company also produces laptop batteries, and claims that the batteries
gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

Step-by-step explanation:

<em>The question is incomplete: To test this claim a sample or population standard deviation is needed.</em>

<em>We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.</em>

<em> NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.</em>

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

The significance level is 0.05.

The sample has a size n=500.

The sample mean is M=3.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{500}}=0.0894

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{3.5-4}{0.0894}=\dfrac{-0.5}{0.0894}=-5.5902

The degrees of freedom for this sample size are:

df=n-1=500-1=499

This test is a left-tailed test, with 499 degrees of freedom and t=-5.5902, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours.

5 0
3 years ago
The interval for this number line is ______.<br><br> A.3<br> B.5<br> C.6<br> D.4
sergejj [24]

Answer:

it is D.4

Step-by-step explanation:

each place on the number line goes up by 4

3 0
3 years ago
Read 2 more answers
​Simplify. ​13(−32.48−14.02)​ Enter your answer, as a decimal to tenths, in the box.
Artyom0805 [142]

If you mean 1/3 the answer is -15.5

1/3(-32.48-14.02) = 1/3(-46.5) = (-15.5)

3 0
3 years ago
-8a-9a+7a what does this equal
allochka39001 [22]
The answer would be -10a
7 0
3 years ago
FIRST TO ANSWER GETS BRAINLIEST: The average age of one hundred Year 9 students is 14 y 10 m. The average age of one hundred Yea
Dvinal [7]

Answer:

2) 14 y 2 m

Step-by-step explanation:

Year 9 students: total age= 100* 14 10/12= 1400 +100*5/6= 1483 y 8 m

Year 8 students: total age= 100* 13 6/12= 1350 y

Total age of 200 students: 1483 y 8 m + 1350 y= 2833 y 8 m

Average age= 2833  8/12  ÷ 200= (12*2833+8)/12 ÷ 200 = 34004/(12*200)= 34004/2400= 14  404/2400 ≈ 14 1/6 y= 14 y 2 m

4 0
3 years ago
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