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3 years ago

Can someone please help me out ASAP I WILL GIVE OUT BRAINLY

Mathematics

2 answers:

Elis [28]3 years ago

8 0

Answer:

x=9

Perimeter:76 ft

Area:342 ft^2

Step-by-step explanation:

8x-54=18

Then solve for x.

Perimeter: add all sides

(2(9)+1)+(2(9)+1)+18+18

Area: multiply length and width

18 x 19 = 342

SashulF [63]3 years ago

3 0

Answer:

x = 9

Perimeter = 74

Area = 342

Step-by-step explanation:

First, find x. to do so, set your formula up as

18 = 8x - 54

18 + 54 = 8x

72 = 8x

72/8 = x

9 = x

Now that you have x, find your missing side

2x + 1 = y, x

2 * 9 + = y

19 = y

Your perimeter is P = 2(l+W)

P = 2(18 * 19)

P = 74

Your area is A = wl

A = 19 * 18

A = 342

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Given, the members of the junior varsity swim team wear glasses is 55%.

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Let the total no. of members be x.

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So, the probability for no. of members not wearing glasses and in 10th grade will be,

$P(E)=\dfrac{0.135x}{x}$

$P(E)=0.135\\\\P(E)=13.5 \%$

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2 years ago

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

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Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

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The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

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Thus our probability becomes

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Part b)

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We have to calculate probability of part 2 as we have already calculated probability of part 1.

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