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3 years ago

6

The distribution of the heights of students in a large class is roughly Normal. Moreover, the mean height is 68 inches, and appr

oximately 95% of the heights are between 62 and 74inches. Thus, the standard deviation of the height distribution is approximately equal to

1 answer:

Thepotemich [5.8K]3 years ago

3 0

Answer:

Probably equal to 6 or 7.

Step-by-step explanation:

You might be interested in

What is a mixed number for 5/2

m_a_m_a [10]

Answer: 2 1/2

Step-by-step explanation: Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction. Hope this helps!! PLZ mark me!!!

8 0

3 years ago

Which of the following terms correctly describe the figure given below? Check all that apply

Evgesh-ka [11]

Answer:

one of the answers is E,F

Step-by-step explanation:

I know this becuase that is a pyramid right there it is an solid but no prims or anything else on there

3 0

3 years ago

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

4 0

3 years ago

the length of a rectangular garden is 3 m greater than the width. The area of the garden is 70m squared. find the dimensions of

Lerok [7]

A = l x w

so you know that the length is 3m longer than the width, so you could use a formula to represent that

w = l + 3

you then substitute the second equation into the first to solve for l

70 = l x (l +3)
70 = l^2 + 3l

you could then rearrange the formula and solve for l using the quadratic formula

0 = l^2 + 3l - 70

l = -3 +- (square root (3)^2 - 4(1)(70)) / 2(1)
l = -3 +- (square root 9 + 280) / 2
l = -3 +- (square root 289) / 2
l = -3 +- 17 / 2

then you solve for the two seperate roots

l = -3 + 17 /2
l = 14 / 2
l = 7

or

l = -3 - 17 / 2
l = -20 / 2
l = -10

since a length cannot be negative, this root is not viable. therefore l = 7

to solve for w you would use

w = l + 3
w = 7 + 3
w = 10

hope this helps! if you did not understand a step or concept please let me know!

4 0

3 years ago

The table shows the relationship

Airida [17]

The linear relationship which exists between the change in weight and number of days of hibernation in a hedgedog can be modeled using the equation y = - 0.03 + 0

Change in weight per day of hibernation = - 0.03
Change in weight for 115 days = - 3.45 ounces

A linear model can be created using the data in the table given using a regression calculator or excel ;

The linear model obtained in the form y = m + bx is :

y = - 0.03x + 0
y = change in weight
x = number of days of hibernation
Intercept = 0
Slope = -0.03

The slope value of the function gives the change in weight value per number of days of hibernation ; which is -0.03.

Using the regression equation, substitute, the vlaue of x = 115
-0.03(115) + 0 = - 3.45

Therefore, the change in weight after 115 days of hibernation is - 3.45 ounces.

Learn more :brainly.com/question/18405415

8 0

2 years ago