How to divide 704 by 3 using long division

Mathematics

1 answer:

shusha [124]2 years ago

3 0

Answer:

Step 1: D for Divide. How many times will 5 go into 65? ...

Step 2: M for Multiply. You multiply your answer from step 1 and your divisor: 1 x 5 = 5. ...

Step 3: S for Subtract. Next you subtract. ...

Step 4: B for Bring down. ...

Step 1: D for Divide. ...

Step 2: M for Multiply. ...

Step 3: S for Subtract.

Step-by-step explanation:

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NEED HELP ASAP! Question in the photo.

Sonja [21]

Answer:

SAS Postulate

Step-by-step explanation:

You can use the SAS (side, angle, side) postulate that says "if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent"

Side AB is proportionate to DE and

Side AC is proportionate to DF.

Angle A and Angle D are the same; and is between the two sides

I hope this helps.

6 0

2 years ago

THANK YOU (PLEASE HELP) <3 C: TYSMMMM PLZZZ Write an equation of a parabola that is wider than the parent function, reflects

My name is Ann [436]

Answer:

y=-2x^2-4

Step-by-step explanation:

The parabola is wider than it's parent function, y = x^2. Therefore the coefficient, of x^2 (when you refer to quadratic equations it's a) should be greater than 1. Let's take 2.

y = 2x^2

Now it is reflected across the x-axis, so the equation turns to y = -2x^2. The vertex is shifted down 4. Note that it is shifted down, along the y-axis. This makes 'c' (again quadratic equations) -4.

Final Equation: y=-2x^2-4

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2 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$