All categories

3 years ago

Pete’s grade on a test was within 5 points of his class average of 94. What is his range of grades on the test?

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

I'm pretty sure it would be 89-99.

irga5000 [103]3 years ago

6 0

Answer: 89 to 99

Step-by-step explanation:

We know that the range of data is given by :-

$\overline{x}\pm E$ , where $\overline{x}$ ios the average of data and E is the margin of error.

Given : Margin of error = $\pm 5$

Class average = $94$

Then, the range of grades on the test will be :-

$94\pm 5=(94-5,\ 94+5)=(89,\ 99)$

Hence, the range of grades on the test = 89 to 99

You might be interested in

Solve for x. a=1/2h(b x)

fredd [130]

A = 1/2 h(b + x)
2a = h(b + x)
2a/h = b + x
x = 2a/h - b

5 0

3 years ago

10ptsplzanswerdizforme

nalin [4]

X=4y+16
y=x/4-4
hope it helps

6 0

3 years ago

Xt-3=y solve the literal equation for the variable T.

faltersainse [42]

Xt-3=y

Move three to the other side as a positive number

xt=3+y

Divide everything by x

t=3+y divided by x

5 0

2 years ago

A bank customer borrowed $4000 for 1 year at 5% simple interest to buy a computer. How much interest did the customer pay

s344n2d4d5 [400]

$4,200 because 1.05 times 4,000 is 4,200

8 0

2 years ago

Differential Equations<br><br> Solve the initial-value problem dydx=e2x/5y4,y(0)=4.<br><br> y(x)=

solniwko [45]

Answer:

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}$

Step-by-step explanation:

$\displaystyle\frac{dy}{dx} = \displaystyle\frac{e^{2x}}{5y^4}$

Cross multiplying, we have,

$5y^4 dy = e^{2x} dx$

Integrating both sides,

$\int 5y^4 dy = \int e^{2x}dx$

We obtain,

$y^5 = \frac{e^{2x}}{2} + C$ , where C is the constant of integration.

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + C }$

We know that y(0) = 4

Putting these value in the above equation, we get C = $\frac{2047}{2}$

Thus,

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}$

5 0

3 years ago