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3 years ago

A test has a mean of 80 and standard deviation of 4. what score would be 1 deviation from the mean

Mathematics

2 answers:

Varvara68 [4.7K]3 years ago

5 0

Answer:

The score from 76 to 84 would be 1 deviation from the mean.

Step-by-step explanation:

It is given that a test has a mean of 80 and standard deviation of 4. It measn

$\mu=80$

$\sigma=4$

We need to find the score that would be 1 deviation from the mean. It means, we need to find the interval $[\mu-\sigma,\mu+\sigma]$ .

$\mu-\sigma=80-4=76$

$\mu+\sigma=80+4=84$

The required interval is [76,84].

Therefore, the score from 76 to 84 would be 1 deviation from the mean.

GarryVolchara [31]3 years ago

4 0

Any score between 76 and 84 will be 1 deviation from the mean.

To figure this out you subtract 4 from 80 (80-4 = 76) and you add 4 to 80 (80+4) = 84 and that tells you that most of the scores will fall between 76 and 84 which is 1 deviation from the mean.

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A bird species in danger of extinction has a population that is decreasing exponentially (A = A0e^kt). Five years ago, the popul

Natalija [7]

Answer:

It'll take approximately 34 years from today.

Step-by-step explanation:

in order to solve this problem we first need to find the rate of change, "k", to do that we will use the given information where the population was 1400 five years ago and its now 1000. Applying this data to the equation gives us:

$A = A_0*e^{k*t}\\1000 = 1400*e^{5*k}\\1400*e^{5*k} = 1000\\e^{5*k} = \frac{1000}{1400}\\ln(e^{5*k}) = ln(\frac{1000}{1400})\\5*k = ln(1000) - ln(1400) \\k = \frac{ln(1000) - ln(1400)}{5} = -0.06729$

We now know the value for "k", we can estimate how many years it will take for the bird population to dip below 100. We have:

$100 = 1000*e^{-0.06729*t}\\e^{-0.06729*t} = \frac{100}{1000}\\ln(e^{-0.06729*t} = \frac{1}{10}\\-0.06729*t = ln(0.1)\\t = -\frac{ln(0.1)}{0.06729} = 34.22$

It'll take approximately 34 years from today.

8 0

4 years ago

9. Calista’s gross annual income is $39,036. She is paid monthly and has 6% deducted from her paychecks for her 403(b). If her e

Delicious77 [7]

Answer:

$203 approx

Explanation:

Gross Annual Income $39036.

Gross Monthly Income would be $\frac{39036}{12}$ i.e $3253 per month.

Out of this gross income, Calista deposits 6% i.e 6 % of $3253 = $195.18

The employer matches it with 4% of the above deduction i.e 4$ contributed for every 100 dollars of employee contribution.i.e 4% of $195.18

= $ 7.8072

Hence the Amount deposited in Calista's retirement plan each month would be: Employee's contribution + Employer's contribution

= $195.18 + $7.8072

= $ 202.9872 or $ 203 approx every month

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3 years ago

Indicate If these triangles are SSS SAS SAA ASA HL or none

Setler79 [48]

In the two triangles, all the three sides are congruent to the corresponding sides of the other triangle. And in such case, the criteria used for congruency is Side Side Side or SSS .

So the correct option is the first option .

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3 years ago

Consider the reduction of the complex figure with a scale factor of 1

snow_tiger [21]

Answer:

Step-by-step explanation:

I dont know how i got it but i tried it and got 100 and my answer was 25

3 0

3 years ago

Read 2 more answers

An SRS of size n was taken to estimate mean body mass index (BMI) for girls between 13 and 19 years of age. The 95% confidence i

hodyreva [135]

Answer:

Option 1) A total of 95% of all teenage girls have body mass index between 19.5 and 26.3.

Step-by-step explanation:

We are given the following in the question:

95% confidence interval of mean body mass index (BMI) for girls between 13 and 19 years of age is:

$(19.5, 26.3)$

Lower limit = 19.5

Upper limit = 26.3

1. A total of 95% of all teenage girls have body mass index between 19.5 and 26.3.

The given statement is false. This is not the right interpretation for confidence interval.

2. The margin of error is 3.4

The confidence interval can be found as

$\text{sample mean}\pm \text{Margin of error}$

Let x be the sample mean and y be the margin of error, thus, we can write

$x - y =19.5\\x + y = 26.3\\2x =45.8 \\x = 22.9\\y = 22.9 - 19.5\\y = 3.4$

Thus, the margin of error is 3.4

Thus, the given statement is true.

3. The value z in the margin of error is 1.96.

The statement is true.

The z-multiplier for a 95% confidence interval used is 1.96

4. A total of 95% of all SRS of size n contain the true mean body mass index

The given statement is true. It is the right interpretation of 95% confidence interval.

6 0

3 years ago