Quares and Differences of Squares

Solve 5y^2+20=80, where y is the real number

soldi70 [24.7K]

Answer:

2√3

Step-by-step explanation:

5y2+20=80

Step 1: Subtract 20 from both sides.

5y^2+20−20=80−20

5y^2=60

Step 2: Divide both sides by 5.

5y^2 /5 = 60/5

y2=12

Step 3: Take square root.

y=√12

y=2√3

3 0

3 years ago

Simplify 7(2x+y)+6(×+5y)

iris [78.8K]

Simplify, 7(2x+y)+6(x+5y).

a: 20x+37y

(7 • (2x + y)) + 6 • (x + 5y)

Step 2 :

Equation at the end of step 2 :

7 • (2x + y) + 6 • (x + 5y)

Step 3 :

Final result :

Thanks,

Deku ❤

4 0

3 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago

Gerald recorded the gains and losses for each play on the football field. Gains were marked with a + and losses with a –. Play 1

Vedmedyk [2.9K]

Answer:

Play 3

Step-by-step explanation:

Given:

$\left\begin{array}{ccccc}{Play}&{1}&{2}&{3}&{4}&{Yards}&{+4}&{-6}&{-2}&{+3}\end{array}\right$

Required [Missing from the question]:

Determine the play with the least change in yard?

To do this, we ignore the sign in front of each yard before we analyze.

So, we have:

Play 1 has a change of 4

Play 2 has 6

Play 3 has 2

Play 4 has 3

From the above analysis, play has the least (which is 2)

4 0

2 years ago

Please help ASAP! what is x=-b/2a?

valina [46]

This is the formula to solve for the vertex.

Example Question:

Find the vertex of y = -0.5x^2 + 100x

-b/2a = -100/2(-0.5) = -100/-1 = 100

The x coordinate of the vertex is 100.

Best of Luck!

P.S. this is 1000th question i've answered :)

5 0

3 years ago