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3 years ago

WILL GIVE BRAINLIEST! 40 POINTS! please answer right and dont be mean

Mathematics

1 answer:

Viefleur [7K]3 years ago

6 0

Answer:

Brainliest are always appreciated :)

Step-by-step explanation:

For Number 1, just make a graph. (THE LINE WILL BE STRAIGHT)

For Number 2:

Proportional is when quantities have the same relative size. In other words they have the same ratio. The cocoa has an equivalent ratio which is why its proportional. This means that as x increases, y increases and as x decreases, y decreases-and that the ratio between them always stays the same.

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Math question down below

Kipish [7]

Answer: 6928

----------------------------------------------------------

Explanation:

We have two areas we need to find: The area of the trapezoid and the area of the rectangle. Let's call these areas A1 and A2.

Area of Trapezoid = (height)*(base1+base2)/2
A1 = h*(b1+b2)/2
A1 = 80*(150+100)/2
A1 = 80*250/2
A1 = 20000
A1 = 10000

Area of Rectangle = (length)*(width)
A2 = L*W
A2 = 48*64
A2 = 3072

Subtract the two areas (A1-A2) to get the difference D
D = A1 - A2
D = 10000 - 3072
D = 6928

This difference D is exactly equal to the shaded area.

3 0

3 years ago

Please help, I dont know how to do this

Mrrafil [7]

Answer:

$\frac{y-2}{4}$

Step-by-step explanation:

Let's start by factoring everything:

$\dfrac{\frac{y^2+y}{y^2-2y}}{\frac{4y+4}{y^2-4y+4}}=$

$\dfrac{\frac{y(y+1)}{y(y-2)}}{\frac{4(y+1)}{(y-2)^2}}=$

Now, you can cancel out some terms:

$\dfrac{\frac{y+1}{y-2}}{\frac{4(y+1)}{(y-2)^2}}=$

Now, when you divide a fraction by another fraction, that is the same as multiplying the first fraction by the reciprocal of the second one:

$\frac{y+1}{y-2}\cdot \frac{(y-2)^2)}{4(y+1)}=$

$\frac{y-2}{4}$

Hope this helps!

4 0

3 years ago

Mapiya writes a series of novels. She earned \$75{,}000$75,000dollar sign, 75, comma, 000 for the first book, and her cumulative

qwelly [4]

Answer:

$E(n)=75000 \times 2^n$

Complete question:

write a function that gives mapiyas cumulative earnings E(n), in dollars when she has written n sequel's

Step-by-step explanation:

According to the question, she earned $75000 for the first book.

Also,We are given that her cumulative earnings double with each sequel that she writes.

Assuming she has written n sequel's

Now since we are given that her cumulative earnings double with each sequel

So, her initial earning will be $2^n$ times

So, her earning will be : $75000 \times 2^n$

Now we are given that cumulative earnings is denoted by E(n)

So, the function becomes : $E(n)=75000 \times 2^n$

Hence a function that gives Mapiya's cumulative earnings E(n), in dollars when she has written n sequel's is $E(n)=75000 \times 2^n$

4 0

3 years ago

Ms. Sunshine is going to bake cookies for her students. She wants each student to receive 2.5 cookies. If she has 25 students in

adell [148]

Answer:

Ms. Sunshine needs to make 62.5 cookies.

Step-by-step explanation:

25 * 2.5 = 62.5

4 0

3 years ago

The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution

tekilochka [14]

Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

Step-by-step explanation:

Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);

$P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0$

$P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693$

P(Xi≥19.0)=0.473

$\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}$

p=0.473

Yi~Ber(0.473)

$P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)$

Then:

$P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0$

$P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)$

$P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5$

Then:

the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

7 0

3 years ago