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3 years ago

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PLEASEEE HELPPP me!!!!!

1 answer:

makvit [3.9K]3 years ago

8 0

Since you have 2/3 and after you give the bus driver chocolate you have 1/3 left, then you have the bus driver 1/3 because 1/3+1/3 equals 2/3. So you give the bus driver 1/3

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Find the midpoint of the line segment between the points (11, -5) and (-3, -7)

alexdok [17]

Answer:

(4, -6)

Step-by-step explanation:

(x1+x2)/2 + (y1+y2)/2 = midpoint

(11+-3)/2 = 8/2 = 4. x = 4

(-5 + -7)/2 = -12/2. y = -6

4, -6 is the midpoint

give brainliest please! hope this helps :)

4 0

2 years ago

The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability

Whitepunk [10]

Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let <em>Y</em> = the time taken to deliver a pizza.

The random variable <em>Y</em> follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

$f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.$

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let <em>X</em> = results of a certain blood test.

It is provided that the random variable <em>X</em> follows a Normal distribution with parameters $\mu = 60$ and $s = 18$ .

The probabilities of a Normal distribution are computed by converting the raw scores to <em>z</em>-scores.

The <em>z</em>-scores follows a Standard normal distribution, N (0, 1).

(a)

Compute the probability that the results are more than 45 as follows:

$P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z$

The percentage of results more than 45 is: $0.7967\times100=79.67\%$

Thus, the percentage of results more than 45 is 79.67%.

(b)

Compute the probability that the results are less than 85 as follows:

$P(X$

The percentage of results less than 85 is: $0.9177\times100=91.77\%$

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

$P(75$

The percentage of results are between 75 and 90 is: $0.1558\times100=15.58\%$

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

$P(20$

Then the probability that the results outside the range 20 to 100 is: $1-0.9736=0.0264$ .

The percentage of results outside the range 20 to 100 is: $0.0264\times100=2.64\%$

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

4 0

3 years ago

Find the unknown side length

ohaa [14]

You use the formula a² +b² =c²
so 4² +b² =8²
solve for b

7 0

3 years ago

An arrangement for one container needs to have at least 4 flowers in it and the total wholesale cost must be less than $12. At w

sesenic [268]

Answer:

the answer is A on edge 2020

Step-by-step explanation:

hope this helped:)

5 0

3 years ago

Read 2 more answers

I really need help I’ve worked it out already though

Vikki [24]

<h2><em>Answer</em><em>:</em></h2><h3><em>7</em><em>5</em><em>(</em><em>5</em><em>r</em><em>)</em><em>+</em><em>7</em><em>5</em><em>(</em><em>3</em><em>d</em><em>)</em><em>+</em><em>7</em><em>5</em><em>(</em><em>4</em><em>f</em><em>)</em></h3>

<em>Solution</em><em>,</em>

<em>Cost </em><em>of </em><em>roses </em><em>in </em><em>1</em><em> </em><em>bouquet=</em><em>5</em><em>r</em>

<em>Cost </em><em>of </em><em>daisies </em><em>in </em><em>1</em><em> </em><em>bouquet=</em><em>3</em><em>d</em>

<em>cost </em><em>of </em><em>green </em><em>filler </em><em>in </em><em>1</em><em> </em><em>bouquet</em><em>=</em><em>4</em><em>f</em>

<em>Total </em><em>no.</em><em> </em><em>of </em><em>bo</em><em>u</em><em>q</em><em>u</em><em>e</em><em>t=</em><em>7</em><em>5</em>

<em>cost </em><em>of </em><em>one </em><em>bouquet=</em><em>5</em><em>r</em><em>+</em><em>3</em><em>d</em><em>+</em><em>4</em><em>f</em>

<em>total=</em><em>7</em><em>5</em><em>(</em><em>5</em><em>r</em><em>+</em><em>3</em><em>d</em><em>+</em><em>4</em><em>f</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>5</em><em>(</em><em>5</em><em>r</em><em>)</em><em>+</em><em>7</em><em>5</em><em>(</em><em>3</em><em>d</em><em>)</em><em>+</em><em>7</em><em>5</em><em>(</em><em>4</em><em>f</em><em>)</em>

<em>hope </em><em>it</em><em> helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

8 0

3 years ago