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3 years ago

Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Mathematics

1 answer:

Vladimir79 [104]3 years ago

4 0

$p\%=\dfrac{p}{100}=p:100$

$1.\ 30\%=\dfrac{30}{100}=\dfrac{3}{10}\\\\30\%=30:100=0.3\\\\2.\ 42\%=\dfrac{42}{100}=\dfrac{42:2}{100:2}=\dfrac{21}{50}\\\\42\%=42:100=0.42\\\\3.\ 18\%=\dfrac{18}{100}=\dfrac{18:2}{100:2}=\dfrac{9}{50}\\\\18\%=18:100=0.18$

$4.\ 35\%=\dfrac{35}{100}=\dfrac{35:5}{100:5}=\dfrac{7}{20}\\\\35\%=35:100=0.35\\\\5.\ 100\%=\dfrac{100}{100}=\dfrac{1}{1}=1\\\\100\%=100:100=1.0\\\\6.\ 29\%=\dfrac{29}{100}\\\\29\%=29:100=0.29$

$7.\ 56\%=\dfrac{56}{100}=\dfrac{56:4}{100:4}=\dfrac{14}{25}\\\\56\%=56:100=0.56\\\\8.\ 66\dfrac{2}{3}\%=\dfrac{66\cdot3+2}{3}\%=\dfrac{200}{3}\%\\\\\dfrac{200}{3}\%=\dfrac{\frac{200}{3}}{100}=\dfrac{200}{3\cdot100}=\dfrac{2}{3}\\\\\dfrac{200}{3}\%=\dfrac{200}{3}:100=66.666...:100=0.666...\approx0.67\\\\9.\ 25\%=\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\\\\25\%=25:100=0.25$

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A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

Standard deviation of defective light bulbs is given by = S.D. = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.67

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3 years ago

There are 84 soccer players in a local league. At tjis rate how many soccer players are there per team?

juin [17]

Answer:

how many teams are there?

Step-by-step explanation:

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2 years ago

G-GPE-S & G-GPE-7: Practice_2

mezya [45]

The required equation is y = -9

Step-by-step explanation:

Step 1 :

Given the line l is perpendicular to the y axis

The equation of the y axis is x = 0

So any line perpendicular to the y axis will have equation as y = k , where k is a constant value for any value of x

Step 2 :

Its given that the line is passing through the point (0.-9). Here the y co ordinate is -9. Hence the perpendicular line has a constant y co ordinate of y = -9 for any value of x

So the required equation is y = -9

Step 3 :

Answer :

The required equation is y = -9

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3 years ago

Solve for the solution: 2m+1≥7

QveST [7]

Answer:

m≥3

Step-by-step explanation:

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