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1 year ago

Solve (6b^2+5)+(8b-20)

Mathematics

1 answer:

Lisa [10]1 year ago

7 0

The sum of the given expression is 6b^2+8b-15

<h3>Sum of expression</h3>

Given the following expression

(6b^2+5)+(8b-20)

We are to take the sum. On expansion;

6b^2+5 + 8b-20

Collect the like terms

6b^2+8b +5 - 20

6b^2+8b-15

Hence the sum of the given expression is 6b^2+8b-15

Learn more on sum of expression here: brainly.com/question/723406

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Answer:

Option (4). 25%

Step-by-step explanation:

The graph attached shows the exponential growth.

Let the graphed function is y = $(1+\frac{r}{100})^{t}$

Here 'r' = Rate of growth

t = Duration of time in years

y = enrollments after time 't' years

Graph shows at time 't' = 0 or initially number of enrollments = 20

After 8 years number of enrollments will be

120 = $20(1+\frac{r}{100})^{8}$

$\frac{120}{20}=(1+\frac{r}{100})^{8}$

6 = $(1+\frac{r}{100})^{8}$

log 6 = $8\text{log}(1+\frac{r}{100})$

0.77815 = $8\text{log}(1+\frac{r}{100})$

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$(1+\frac{r}{100})=1.251$

$\frac{r}{100}=0.251$

r = 25.1%

r ≈ 25%

Therefore, Option (4) will be the answer.

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3 years ago

BRAINLIEST!!!<br> 18. Point p is chosen at random on EH. Find the probability that p is on FG.

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Answer:

$\frac{2}{5}$

Step-by-step explanation:

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This means that the probability of p being on FG would be $\frac{6}{15}$ which can be simplified to $\frac{2}{5}$

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2 years ago

Read 2 more answers

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Answer:

{GT, GH, BT, BH, RT, RH};

1 /5;

Mutually exclusive ;

Not mutually exclusive.

Step-by-step explanation:

Given :

Green cards, G = 4

Blue cards, B = 4

Red cards, R = 2

For a coin toss :

{H, T}

A card is picked, then a coin is tossed :

Sample space :

{GT, GH, BT, BH, RT, RH}

2.) probability of picking a green card, then probability of landing a head on a coin toss

P(A) = number of required outcome / Total possible outcomes

P(A) = P(green card) * P(Head)

P(A) = (4 / 10) * (1/2)

P(A) = 4 /20

P(A) = 1/5

C.)

Both A and B are mutually exclusive, since both event A and event B cannot occur together, since both red and green cannot be picked during a single pick, this either a red is picked or green is picked, then they are A and B are mutually exclusive.

D.) Event A and C are not mutually exclusive, picking a green card, event A and picking a red or blue card, event B. Both event can happen simultaneously, hence, event A and B are not mutually exclusive.

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2 years ago

Alyssa took a math test for 20 questions and she answered 16 questions correctly in order to get an a she needs to answer at lea

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Answer:

She did not qualify the exam.

Step-by-step explanation:

Given data

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Percentage of question answered correctly

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But in order to get in the exam the student must have to score at least 90 %. But she got only 80 % so she did not qualify the exam.

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2 years ago