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3 years ago

15

Dr. Ibanez recorded the amount of snowfall, in centimeters, at 10 different locations during 1 day and then arranged the amounts

from smallest to largest. The amounts were 3, 3, 3, 3, 4, 4, 4, 5, 5, and 6 cm. The mean of these measurements is 4 cm.
What is the mean absolute deviation?

A.
0 cm

B.
0.8 cm

C.
1 cm

D.
2.5 cm

2 answers:

iren2701 [21]3 years ago

7 0

Answer: the answer is b 0.8

Step-by-step explanation:

Mkey [24]3 years ago

6 0

<span>So we want to know what MAD from this set of data is that has the mean value of 4. First we determine the distance d of each number from the set from the mean value: d1=4-3=1, d2=4-4=0, d3=5-4=1, d4=6-4=2. So our set of distances is {1,1,1,1,0,0,0,1,1,2}. Now we can determine MAD: MAD=(1+1+1+1+0+0+0+1+1+2)/10=0.8.</span>

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a positive number is 7 times another number if 3 is added to both the numbers then one of the new number becomes 5 by 2 times th

Neko [114]

Answer:

7 and 1

Step-by-step explanation:

Let the numbers be a and b.

<u>A positive number is 7 times another number:</u>

a = 7b

<u>If 3 is added to both the numbers then one of the new number becomes 5 by 2 times the other new number:</u>

a+3 = 5/2 × (b +3)

<u>To solve this we substitute </u><u>a</u><u> with </u><u>7b</u><u> in the second equation:</u>

7b + 3 = 5/2 × (b +3) ⇒ multiplying both sides by 2
14b + 6 = 5b + 15 ⇒ collecting like terms
14b - 5b = 15 - 6
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a= 7b
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<u>So the numbers are</u> 7 and 1

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3 years ago

Which shows 0.23 repeated as a fraction.<br> A. 2/33 <br> B. 7/33 <br> C. 23/99 <br> D. 7/30

sergij07 [2.7K]

Answer: c

Step-by-step explanation: 23/99 as a fraction

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3 years ago

Is 5- bigger than 9-? can you please tell be why

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2 years ago

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The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

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zavuch27 [327]

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