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3 years ago

Find 120% of 18. a) 22.5 b) 23 c) 15 d) 21.6

Mathematics

1 answer:

Zina [86]3 years ago

7 0

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25-20x+4x² heh help pls and hurry :((

yulyashka [42]

25- 20x + 4x^2
=4x^2 - 20x + 25
=4x^2 - x(10+10) +25
=4x^2 - 10x - 10x + 25
=2x(2x - 5) - 5( 2x - 5)
=(2x - 5)(2x - 5)
=(2x - 5)^2

6 0

3 years ago

Read 2 more answers

A bag contains 4 b l u e marbles, 3 w h i t e marbles, and 10 r e d marbles. If a marble is drawn from the bag, replaced, and an

lawyer [7]

Answer:

40/289

Step-by-step explanation:

We can see these two events are independent as Regina replaced marble after drawing one marble.

We can find possible outcomes by adding number of blue, red and yellow marbles.

possible outcomes= 4+3+10= 17

Let us find probability of getting a blue marble on first draw.

$P( blue\ marble)= \frac{4}{4+3+10} = \frac{4}{17}$

Now we will find probability of getting red marble.

$P(red/marble)= \frac{10}{17}$

We can find probability of getting a blue marble and then red marble by multiplying both probabilities.

$P(\text{ Blue then red})=\frac{4}{17}* \frac{10}{17} =\frac{40}{289}$

4 0

4 years ago

A student wants to use the property

Ratling [72]

Answer:

hyoyr nhggggfryuvsrthjugdhba jgrybcwqetjkoplhsaxvnnutfvhyu

6 0

3 years ago

3.14 and

stepladder [879]

Answer:

471ft^3

Step-by-step explanation:

5^2 x 6 x 3.14

25 x 6 = 150

150 x 3.14 = 471

5 0

3 years ago

According to a study conducted by the Gallup Organization, the the proportion of Americans who are afraid to fly is 0.10. A rand

notsponge [240]

Answer: 0.1457

Step-by-step explanation:

Let p be the population proportion.

Given: The proportion of Americans who are afraid to fly is 0.10.

i.e. p= 0.10

Sample size : n= 1100

Sample proportion of Americans who are afraid to fly = $\hat{p}=\dfrac{121}{1100}=0.11$

We assume that the population is normally distributed

Now, the probability that the sample proportion is more than 0.11:

$P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]$

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

3 0

3 years ago