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2 years ago

A car drives at a rate of 65 mi/hr. what is the speed of the car in ft/s? note: 1 mi = 5280 ft

Mathematics

2 answers:

ANTONII [103]2 years ago

6 0

Answer:. 95.33 ft/sec

Multiply 65 by 5280 to get 343200 feet per hour. Then Divide by 3600. There are 60 minutes in an hour, and 60 seconds in a minute.

343200 ÷ 3600 = 95.333

lana66690 [7]2 years ago

5 0

Answer:. 95.33 ft/sec

Multiply 65 by 5280 to get 343200 feet per hour. Then Divide by 3600. There are 60 minutes in an hour, and 60 seconds in a minute.

343200 ÷ 3600 = 95.333

Step-by-step explanation:

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Answer:

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3 years ago

Read 2 more answers

The graph shows a probability distribution. What is P(1.5 ≤ X ≤ 3)? Enter your answer, as a decimal, in the box.

Bumek [7]

The probability from 1.5 ≤ x ≤ 3 can be calculated by dividing the Area from x=1.5 to x=3 by the total Area of the distribution.

The given distribution is rectangular shaped, so its Area will be = Length x Width = 1 x 3 = 3 square units

From x = 1.5 to x = 3, the length is 1.5 and width is 1. So the area between these two intervals = 1.5 square units.

Thus, <span>P(1.5 ≤ X ≤ 3) = 1.5/3 = 0.5
</span>

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3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .