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2 years ago

1) Blair has a box of candy to sell. She has 4 chocolate bars, 4 fruit chews, 1 pack of

Mathematics

2 answers:

blsea [12.9K]2 years ago

7 0

Answer:

12 are the possible outcomes

Step-by-step explanation:

because probability is a game of chance

possible outcomes/ total outcomes

GuDViN [60]2 years ago

5 0

Answer:

Step-by-step explanation:

If she's only selecting one candy, then you simply need to count how many candies there are.

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Define equivalent expressions. Give an example of two equivalent expressions.

Tasya [4]

Answer:

Hello There!

Step-by-step explanation:

Equivalent expressions are expressions that work the same even though they look different.

Examples of two equivalent expressions are:

-4(x + 2) and 4x + 8

-2y+5y−5+8 and 7y+3

These are equivalent expressions as they have the same value.

hope this helps,have a great day!!

~Pinky~

3 0

3 years ago

Pls help i don’t know how to do this

algol [13]

Answer: $\displaystyle\\\frac{2\pi }{3},\ \frac{5\pi }{3} .$

Step-by-step explanation:

$\displaystyle\\tan\theta=-\sqrt{3}\ \ \ \ 0\leq \theta\leq 2\pi \\\theta =\frac{2\pi }{3}+\pi N\ \ \ (N=0,\ 1,\ 2,\ 3\ ...)\\ N=0\\\theta=\frac{2\pi }{3} +\pi *0\\\theta=\frac{2\pi }{3}\ \ \ \ ( 0\leq \theta\leq 2\pi)\\ N=1\\\theta=\frac{2\pi }{3}+\pi *1 \\\theta=\frac{2\pi }{3} +\pi \\\theta=\frac{2\pi +3\pi }{3} \\\theta=\frac{5\pi }{3} \ \ \ ( 0\leq \theta\leq 2\pi)\\$

$N=2\\\theta=\frac{2\pi }{3}+2\pi \\ \theta=\frac{2\pi +3*2\pi }{3}\\ \theta=\frac{2\pi +6\pi }{3} \\\theta=\frac{8\pi }{3} \\\theta=2\frac{2}{3} \pi \ \ \ \ (\notin0\leq \theta\leq 2\pi).\\$

6 0

1 year ago

X-3<-12 Choose the correct graph for the inequality?

olga55 [171]

The correct graph is the last one
x -3<-12
x<-12+3
x<-9

7 0

3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

$H_0$ : The true average penetration is 50 mils

$H_a$ : The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the level of confidence is α = 0.05 then the $z_\alpha$ score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)