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3 years ago

Math 7 Grade 7

1 answer:

5 0

The first is independent of one another because the probabilities of picking the marbles doesnt change because the first marble was placed back in. Now the second one is dependent because the first card picked wasnt placed back which changes the probability of the rest of the cards.

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A 15 foot ladder is leaning against a wall. if the top slips down the wall at a rate of 4 ft/s, how fast will the foot be moving

hoa [83]

<span>From the pythagorean theorem, we know that x^2 and y^2 = 15^2, where x and y, where x is the base moving from the wall and y is the distance of the ladder sliding down. This means that y = sqrt(225 - x^2). Plugging in x = 13 gives us y = 2sqrt(14). Differentiating both sides of the original equations gives us dy/dt = 13 x 4 / 2sqrt(14) = 52/2sqrt(14) = 26/sqrt(14) ft/s.</span>

7 0

3 years ago

How do I find the quotient 7/8÷2/9

Lena [83]

$\dfrac{\frac{7}{8}}{\frac{2}{9}}=\frac{7}{8}\cdot\frac{9}{2}=\frac{63}{16}$

5 0

3 years ago

Read 2 more answers

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

3 0

3 years ago

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

german

For this case we have the following function:
$f(x) = tan(\frac{x^2}{9})$
The average rate of change is given by:
$AVR = \frac{f(x2) - f(x1)}{x2- x1}$
Evaluating the function for the given interval we have:
For x = 1.25:
$f(1.25) = tan(\frac{1.25^2}{9})$
$f(1.25) = 0.18$
For x = 2:
$f(2) = tan(\frac{2^2}{9})$
$f(2) = 0.48$
Then, replacing values we have:
$AVR = \frac{0.48 - 0.18}{2 - 1.25}$
$AVR = 0.4$
Answer:
the average value of on the closed interval [1.25, 2] is:
$AVR = 0.4$

3 0

3 years ago

Complete the problem

nika2105 [10]

Answer:

The answer is C..........

4 0

3 years ago