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3 years ago

11

Please help as soon as possible

2 answers:

zaharov [31]3 years ago

7 0

24 students. you would multiply 9 by the reciprocal of 3/8

crimeas [40]3 years ago

7 0

I think there is 27 students in math class for number 3

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8 scholars were sharing 6 brownies each scholar got an equal amount how much did each scholar get

Eva8 [605]

Answer:

4/3 or 1 1/3 brownie

Step-by-step explanation:

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4 years ago

Use the Empirical Rule to answer each question.

Oliga [24]

Step-by-step explanation:

(a)

the probability of parcels weighing 35oz or less is

0.841

the probability of parcels weighing 14oz or less is

0.023

the probability of parcels weighing between 14 and 35oz is the probabilty of 35oz minus the probability of 14oz

0.841 - 0.023 = 0.818

the percentage is the probability × 100 = 81.8%

(b)

the probability of parcels weighing more than 49oz is 1 minus the probability to weigh less than 49oz.

the probability to weigh less than 49oz is

0.99865

so, the probability to weigh more than 49oz is

1 - 0.99865 = 0.00135

the percentage is again probability×100 = 0.14%

7 0

3 years ago

Name the property of equality that justifies going from 3x+6x to (3+6)x

marta [7]

The distributive property

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3 years ago

Solve the equation by using the basic properties of logarithms.

Tom [10]

Answer:

B is the awnser

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago