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2 years ago

PLEASE HELP. I HAVE LIMITED TIME.

Mathematics

1 answer:

Lana71 [14]2 years ago

3 0

I think this is it, but i’m not sure about the last one

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Hey can you help me fast!!!

scZoUnD [109]

Answer:

Area= 4 ft²

Step-by-step explanation:

Area= (2)(2)

multiply 2 × 2= 4

Area= 4 ft²

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hope it helps...

have a great day!!

4 0

2 years ago

What would the equation be for the points (0,-2) and (2,6) be? (SLOPE INTERCEPT FORM y=mx+b)

Paladinen [302]

Answer:

Slope m=4

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

We have A(0,-2)->x1=0, y1=-2

B(2,6)->x2=2, y2=6

7 0

2 years ago

The coefficient of x in -7xy is -7

Oksanka [162]

Answer:

True: the leading coefficient in the monomial is -7

Step-by-step explanation:

Math

5 0

2 years ago

Find the dimensions of a right-circular cylinder that is open on the top and closed on the bottom, so that the can holds 1 liter

anzhelika [568]

Volume of cylinder:
V = πr²h
The desired volume is 1 Liter = 1000 cm³
1000 = πr²h
h = 1000/πr²

Surface area of cylinder:
S.A = 2πr² + 2πr²h
We substitute the value of h from the first equation:
S.A = 2πr² + 2πr(1/πr²)
S.A = 2πr² + 2/r
Now, to minimize surface area, we differentiate the expression with respect to r and equate to 0.
0 = 4πr - 1000/r²
4πr³ - 1000 = 0
r = 4.3 cm
h = 17.2 cm

5 0

3 years ago

The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes

Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(completing the exam in one hour or less)

P(x < 60)

$P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 60) =0.023= 2.3\%$

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

$P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%$