Which fraction is equivalent to <img src="https://tex.z-dn.net/?f=%5Cfrac%7B25%7D%7B35%7D" id="TexFormula1" title="\frac{25}{35}

The circle below is centered at the origin and has a radius of 4. What is its equation?

Alecsey [184]

Answer:

C) x^2 + y^2 = 16

Step-by-step explanation:

<h2><u>Find the equation </u></h2>

since the radius is 4 this means that anywhere from the center of the circle to the edge of the circle is 4. hence :

place the horizontal line of the radius where y = 0 .

this means that if y = 0 , then x = 4

therefore :

x^2 + y^2

4^2 + 0^2

= 16

<h3><u>hence the answer is C ) x^2 + y^2 = 16</u></h3>

7 0

3 years ago

At Sami's Shoe Warehouse, it takes fraction 4 over 5 of a day to complete fraction 1 over 10 of an order of sneakers. At this ra

bogdanovich [222]

B.) 8 days is the answer.

5 0

4 years ago

Read 2 more answers

a)b 256What are the values of a and b in the equation(4x[4x2- ?○ a-2, b-4○ a=-2, b-4○ a-2, b=-4○ a-2, b=-422aaaa0000C日

klasskru [66]

Hi.... your answer is below:

$\left( 4.x^{a} \right)^{b}=\dfrac{256}{x^{8}}\rightarrow 4^{b}x^{ab}=256x^{-8}\rightarrow 2^{2b}.\,x^{ab}=2^{8}.\,x^{-8}\rightarrow\\ \text{by comparison}\\ 2^{2b}=2^{8}\rightarrow 2b=8\rightarrow \boxed{b=4}\\ x^{ab}=x^{-8}\rightarrow x^{4a}=x^{-8}\rightarrow 4a=-8\rightarrow \boxed{a=-2}$

Thanks....

8 0

4 years ago

An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the

neonofarm [45]

Answer:

There is a 33.10% probability that there is at least one defective integrated circuit.

Step-by-step explanation:

For either integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

The electonic product contains 40 integrated circuits, so $n = 40$ .

The probability that any integrated circuit is defective is 0.01, so $\pi = 0.01$

What is the probability that there is at least one defective integrated circuit?

Either there is at least one defective integrated circuit, that is probability $P(X > 0)$ , or there are no defective integrated circuits, that is probability $P(X = 0)$ . The sum of these probabilities is decimal 1. We want to find $P(X>0)$ .

$P(X > 0) + P(X = 0) = 1$

$P(X > 0) = 1 - P(X = 0)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{40,0}.(0.01)^{0}.(0.99)^{40} = 0.6690$

$P(X > 0) = 1 - P(X = 0) = 1 - 0.6690 = 0.3310$

There is a 33.10% probability that there is at least one defective integrated circuit.

3 0

3 years ago

Read 2 more answers

7. Consider the purchase of two cereal boxes.

AveGali [126]

Answer:

a) 0.25

b) 0.25

c) 0.0625

Step-by-step explanation:

The complete question is:

Do you remember when breakfast cereal companies placed prizes in boxes of cereal? Possibly you recall that when a certain prize or toy was particularly special to children, it increased their interest in trying to get that toy. How many boxes of cereal would a customer have to buy to get that toy? Companies used this strategy to sell their cereal.

One of these companies put one of the following toys in its cereal boxes: a block (B), a toy watch (W), a toy ring (R), and a toy airplane (A). A machine that placed the toy in the box was programmed to select a toy by drawing a random number of 1 to 4. If a 1 was selected, the block (or B) was placed in the box; if a 2 was selected, a watch (or W) was placed in the box; if a 3 was selected, a ring (or R) was placed in the box; and if a 4 was selected, an airplane (or A) was placed in the box. When this promotion was launched, young children were especially interested in getting the toy airplane.

What is the probability of getting an airplane in the first cereal box?

Since the machine randomly selects toys, each toy has the same probability of being obtained in a cereal box.

Then, the total outcomes are 4 and the probability of getting an airplane in the first cereal box is 0.25 (25%).

What is the probability of getting an airplane in the second cereal box?

Two independent events do not change the probability of occurrence of one event or another.

The probability of getting an airplane in the second cereal box is 0.25 (25%).

What is the probability of getting airplanes in both cereal boxes?

P(1°∩2°)= P(1°) × P(2°) = $\frac{1}{4} \times \frac{1}{4} =\frac{1}{16}$

P(1°∩2°)= 0.0625 = 6.25%

4 0

3 years ago