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Mathematics

1 answer:

Sphinxa [80]3 years ago

7 0

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What is the maximum area for a rectangle whose perimeter is 48 meters?

MakcuM [25]

Answer:

12m²

Step-by-step explanation:

For a rectangle, with length L and width W,

the perimeter is given as

Perimeter,

P = (2 x Length) + (2 x Width)

P = 2L + 2W

It is given that the perimeter is 48, hence

48 = 2L + 2W (divide both sides by 2)

24 = L + W

L = 24 - W -----> eq 1

Also realize that the Area of a Rectangle is given by

A = L x W -----> eq 2

Substituting eq 1 into eq 2,

A = (24 - W) x W

A = -W² + 24W

Recall that for a quadratic equation y = ax² + bx + c, the maxima or minima is given by y(max) = -b/2a

In this case, b = 24 and a = -1

-b/2a = -24/[ 2(-1) ] = 12

Hence for A to be maximum A(max) = 12m² (Answer)

5 0

3 years ago

Breyers is a major producer of ice cream and would like to test the hypothesis that the average American consumes more than 17 o

zhenek [66]

Answer:

The p-value would be 0.0939

Step-by-step explanation:

We can set a standard t-test for the Null Hypothesis that $H_0: 18.2\geq 17$

The test statistic then takes the form

$t=\frac{18.2-17}{3.9/\sqrt{15}}=1.317$

with this value we then can calculate the probability that is left to the right of this value $P(X>1.317)$ . From theory we know that t follows a standard normal distribution. Then $P(X>1.317)=0.0939$ which is smaller than the p-value set by Breyers of 0.10