Determine whether the ratios are proportional:

Kendra has 3/4 of a bag of hair ribbons. She wants to share so that each of her friends gets 3/8 of the box. How many friends ca

aalyn [17]

3/4 = 6/8 and 3/8 is half of 6/8. If she wanted to give each friend 3/8 of the bag, she could give a portion to only 2 friends.

7 0

4 years ago

Find the equation of the exponential function represented by the table below

Softa [21]

Answer:

$y=4(4)^x$

Step-by-step explanation:

Exponential functions are represented in the form $y=a(b)^x$ , where a is the initial value and b is the multiplier.

Here, we see that when x = 0, y = 4. This means that 4 is our initial value, or a. You can look at the a variable almost being like the y-intercept of a graph.

Next, we must solve for b. We can do 16 ÷ 4 to find the slope, or multiplier of this function, because 16 is 1 input x-value up from 0. 16 ÷ 4 = 4.

So, we now have all the needed variables, so we plug them into the formula

$y=a(b)^x$

8 0

3 years ago

What is /12x^8| /3x2 in simplest form where x>0

svetoff [14.1K]

Answer:

2x^3

C

Step-by-step explanation:

6 0

2 years ago

Please help people this is my assignmentt

sertanlavr [38]

Answer:

its area is 22 square unit

3 0

1 year ago

Read 2 more answers

Over 500 million tweets are sent per day (digital Marketing ramblings website, December 15, 2014). Assume that the number of twe

Aleksandr [31]

Answer:

(a) The probability that Bob receives no tweets during his lunch hour is 0.0002.

(b) The probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c) The expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d) The probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

Step-by-step explanation:

Let X = number of tweets.

The random variable X follows a Poisson distribution with parameter λ = 7.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-7}(7)^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the probability that Bob receives no tweets during his lunch hour as follows:

$P(X=0)=\frac{e^{-7}(7)^{0}}{0!}\\=\frac{0.000192\times1}{1} \\=0.000192\\\approx0.0002$

Thus, the probability that Bob receives no tweets during his lunch hour is 0.0002.

(b)

Compute the probability that Bob receives at least 4 tweets during his lunch hour as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-7}(7)^{0}}{0!}-\frac{e^{-7}(7)^{1}}{1!}-\frac{e^{-7}(7)^{2}}{2!} - \frac{e^{-7}(7)^{3}}{3!}\\=1-0.0002-0.0064-0.0223-0.0521\\=0.9190$

Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c)

The average number of tweets in 60 minutes is 7.

Then the average number of tweets in 1 minute is, $\frac{7}{60}$ .

Hence, the average number of tweets during 30 minutes is, $\frac{7}{60}\times30=3.5$

Thus, the expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d)

Compute the probability that Bob receives no tweets during the first 30 minutes of his lunch hour as follows:

$P(No\ tweets) = \frac{e^{-3.5}(3.5)^{0}}{0!}=\frac{0.0302\times1}{1} =0.0302$

Thus, the probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

3 0

4 years ago