All categories

2 years ago

98 is the product of Chrissy's height and 7 how do I make this a equation

2 answers:

8 0

It's not that hard, It's all substitution! (^O^)/ We would have to assign letters with the key information, soo... H - height C - Chrissy The formula would be: C = 98h In word form; Chrissy equals 98 in height Hope it helps!=^•^=

In-s [12.5K]2 years ago

8 0

Understanding the language in math product means to multiply and means plus, therefore h= Chrissy's height 98=h (h+7) 98=h^2+7h o= h^2+7h-98 factor (h-7). (h+14)=0 h= (7 , -14)

You might be interested in

How do I find the slope and y intercept of y=4x-4

Scorpion4ik [409]

Answer:

the slope is 4/1 because it's 4x and the y intercept is -4

8 0

2 years ago

Read 2 more answers

Which of these equations has a graph that is tangent to the x-axis at one of its intercepts? Choose all the polynomials that hav

Verdich [7]

Answer:

y = ( x + 2)( x + 6)

Step-by-step explanation:

4 0

3 years ago

What is the absolute value of the difference in the x-values between (1,7) and (3,11)

Marysya12 [62]

Answer:

Step-by-step explanation:

The first numbers in the ordered pairs are the x values;

Difference = 1 - 3 = -1

The absolute value is 2.

7 0

2 years ago

Stuck on this problem of my hw. could use some help!

Mazyrski [523]

Answer:

12 is the answer

Step-by-step explanation:

4 0

2 years ago

A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x < 0 and p(x) = 1 for x &gt

34kurt

$F_X(x)=\begin{cases}0&\text{for }x2\end{cases}$

Recall that the PDF is given by the derivative of the CDF:

$f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}$

The mean is given by

$\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23$

The median is the number $M$ such that $F_X(x)=\mathbb P(X\le M)=\dfrac12$ . We have

$F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2$

but both roots can't be medians. As a matter of fact, the median must satisfy $0\le M\le2$ , so we take the solution with the negative root. So $M=2-\sqrt2$ is the median.

3 0

3 years ago