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2 years ago

Help how can I Write the given standard equation of the quadratic in intercept form

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

7 0

Answer:

f(x) = (x - 10) (x + 10)

Step-by-step explanation:

The equation looks like the difference of squares so 5 and 20 are out

-10 x -10 = 100

-10 x 10 = - 100

so (x - 10) (x + 10)

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Name the value of each 2in 222,222

mixas84 [53]

Ones,tenths,hundreths, thousands,tenthousands,hundredthousands

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2 years ago

In January, a dealership sold 164 vehicles. In February, the same dealership sold 23.8% fewer vehicles than they sold in January

sattari [20]

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2 years ago

6. As part of their employee benefits, all workers at Light and Power Electric Company

Semenov [28]

Answer:

$1769.06

Step-by-step explanation:

The sum of Mia's three highest salaries is ...

$92,000 +94,800 +96,250 = $283,050

so the average is ...

$283,050/3 = $94,350

Her pension is then ...

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_____

<em>Additional comment</em>

This is equivalent to the total of those salaries, divided by 160.

T/3 × 0.01875 = T × (0.01875/3) = T × 0.00625 = T/160

Sometimes, there are simpler ways to calculate things like this.

8 0

2 years ago

Which sum will be neither positive nor negative. a. 3+2 b. -4+(-4) c. -6+6 d. 7+(-6)

anastassius [24]

Answer:

Step-by-step explanation:

6 0

2 years ago

A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey

Lubov Fominskaja [6]

Answer:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

Step-by-step explanation:

For this case we have the following distribution given:

X 0 1 2 3 4 5 6

P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02

For this case we need to find first the expected value given by:

$E(X) = \sum_{i=1}^n X_i P(X_I)$

And replacing we got:

$E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61$

Now we can find the second moment given by:

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97$

And the variance would be given by:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

8 0

3 years ago