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3 years ago

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There are 13 boys and 15 girls in Ms. Green’s 3rd period math class. Five girls and four boys have already been called on today.

If Ms. Green does not call on the same student twice, what is the probability the next student she calls on will be a girl?

1 answer:

katrin2010 [14]3 years ago

6 0

The probability that the teacher will choose a girl would be 10/9. I it needs to be simplified it would be 1 1/9

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Jacques deposited $1900 into an account that earns 4% interest. After t year, jaques has $3875.79in the account. How long was th

Sladkaya [172]

Deposit = $1900
Amount After interest = $3875.79
% interest = 4%
Interest = 1975.79

Time For Interest Period

4/100 × 1900 × t = 1975.79
76 t = 1975.79
=26 years

Answer = 26years

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3 years ago

1. Solve the given linear equations for 'x':

postnew [5]

i ) 3(2x - 3) = 4(2x + 4)

$6x - 9 = 8x +1 6 \\ \\ 6x - 8x = 16 + 9 \\ \\ - 2x = 25 \\ \\ x = - \frac{25}{2} .$

ii )2(x + 2) + 5(x + 5) = 4(x-8) + 2(x - 2)

$2x + 4 + 5x + 25 = 4x - 32 + 2x - 4 \\ \\ 7x + 29 = 6x - 36 \\ \\ 7x - 6x = - 36 - 29 \\ \\ x = - 65.$

<h2>----------------------</h2>

2. 3x /4 - 7 /4 = 5x + 12

$\frac{3x}{4} - \frac{7}{4} = 5x + 12 \\ \\ \frac{3x}{4} - 5x = 12 + \frac{7}{4} \\ \\ \frac{3x - 20x}{4} = \frac{48 + 7}{4} \\ \\ \frac{ - 17x}{4} = \frac{55}{4}$

cancel 4 from both sides

$- 17x = 55 \\ \\ x = \frac{ - 17}{55} .$

<h2>-------------------</h2>

3. x +7 - 8x/3 = 17/6 - 5x/8

$x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{8}$

put x = 4

$4 + 7 - \frac{ 8\times 4}{3} = \frac{17}{6} - \frac{5 \times 4}{8} \\ \\ 11 - \frac{32}{3} = \frac{17}{6} - \frac{20}{8} \\ \\ \frac{33 - 32}{3} = \frac{68 - 60}{24} \\ \\ \frac{1}{3} = \frac{8}{24} \\ \\ \frac{1}{3} = \frac{1}{3} .$

L.H.S. = R.H.S.

<h2>--------------------</h2>

4. Let 3 consecutive integers : x , x+1 , x+2

As per the question : 2x + 3(x+1) + 4(x+2) = 56

9x + 11 = 56

Transpose 11 to RHS

9x = 45

Divide both sides by 9

x = 5 ,

x + 1 = 5 + 1 = 6

x + 2 = 5 + 2 = 7

Numbers are 5 , 6 and 7 .

<h2>-----------------------</h2>

5. Let the age of Jane’s younger sister is x.

Age of Jane will be x + 6

As per the question, after 10 years the sum of their ages will be 50.

After 10 years,

age of Jane = x + 16

age of her younger sister = x + 10

Therefore,

x + 16 + x +10 = 50

2x + 26 = 50

2x = 24

x = 12

Thus, the present age of Jane’s younger sister is 12 years.

And of Jane’s is 12 + 6 = 18 years.

<h2>---------------------</h2>

6. Let the breadth of rectangular swimming pool be x metres.

Then, its length = (2x + 2) metres.

We know that the Perimeter of a rectangle

= 2(Length + Breadth)

= 2(2x + 2 + x) = 6x + 4

According to the given condition, we have

6x + 4 = 154

6x = 154 − 4

6x = 150

[Dividing both sides by 6]

x = 25

Thus, breadth = 25 m and

length = 25 × 2 + 2 = 52 m .

<h2>---------------------</h2>

7. Let the digit in ten's .

Place be x and the digit in the one's place be y.

x − y = 5 ⟶ (i)

Two digit number = 10x + y

Two digit after reversing the digits = 10y + x

According to question ,

10x + y + 10y + x = 99

11x + 11y = 99

x + y = 9 ⟶ (ii)

On adding (i) and (ii),

2x = 14

x = 7

Putting the value of x in equation (i)

x − y = 5

y = 2

Number = 10x + y

=72 .

<h2>-----------------------</h2>

8. Let the speed of the steamer in still water be x km/h.

Then , the speed downstream = (x+2) km/h

and the speed upstream = (x−2) km/h

Given , distance covered in 4 hours downstream = distance covered in 5 hours upstream

4(x + 2) = 5(x − 2)

4x + 8 = 5x − 10

4x − 5x = − 10 − 8

[Transposing 5x to LHS and 8 to RHS]

−x = −18 or x = 18 km/h .

<h2>-----------------------</h2>

9. Let the number of notes of different denomination be x.

∴ Numbers of Rs 100 notes , 50 notes and Rs 10 notes will be 2x , 3x and 5x respectively.

Amount of 100 Rs notes ⇒Rs 100 × 2x = 200x

Amount of 50 Rs notes ⇒Rs 50 × 3x = 150x

Amount of 10 Rs notes ⇒Rs 10 × 5x = 50x

As per the question

Total amount = 400000

200x + 150x + 50x = 400000

400x = 400000 , x = 1000

No of Rs 100 notes = 2x = 2 × 1000 = 2000

No of Rs 50 notes = 3x = 3 × 1000 = 3000

No of Rs 10 notes = 5x = 5 × 1000 = 5000 .

<h2>------------------------</h2>

10. Let the age of the man be x

Then age of his son becomes (x − 24)

2 years later from now,

Age of man will be = x + 2

and age of his son will be = (x − 24 + 2) = x − 22

According to question,

2(x − 22) = x + 2

i.e., 2x − 44 = x + 2

i.e., x = 46

Therefore ,

Present age of man = 46 years

And present age of his son = 46 − 24 = 22 years .

4 0

2 years ago

Show how to rewrite 1/5 with the denominator 10.

Gnoma [55]

Answer:

first blank: 2

second blank: 2

8 0

3 years ago

The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party,

Mama L [17]

Answer:

$(a)\ P(x = 0) = 0.2725$

$(b)\ P(x \ge 1) =0.7275$

$(c)\ P(x \le 2) = 0.8948$

Step-by-step explanation:

Given

$n = 8$ --- 8 friends

$p = 15\%$ --- proportion that one-time fling

This question is an illustration of binomial probability, and it is represented as:

$P(X = x) = ^nC_x* p^x * (1 - p)^{n-x}$

Solving (a): P(x = 0) --- None has done one time fling

$P(x = 0) = ^8C_0* (15\%)^0 * (1 - 15\%)^{8-0}$

$P(x = 0) = 1* 1 * (1 - 0.15)^{8}$

$P(x = 0) = 0.85^8$

$P(x = 0) = 0.2725$

Solving (b): $P(x \ge 1)$

To do this, we make use of compliment rule:

$P(x = 0) + P(x \ge 1) =1$

Rewrite as:

$P(x \ge 1) =1 - P(x = 0)$

$P(x \ge 1) =1 - 0.2725$

$P(x \ge 1) =0.7275$

Solving (c): $P(x\le 2)$ --- Not more than 2 has one time fling

This is calculated as:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

We have:

$P(x = 0) = 0.2725$

$P(x = 1) = ^8C_1* (15\%)^1 * (1 - 15\%)^{8-1}$

$P(x = 1) = 8* (0.15) * (1 - 0.15)^7$

$P(x = 1) = 0.3847$

$P(x = 2) = ^8C_2* (15\%)^2 * (1 - 15\%)^{8-2}$

$P(x = 2) = 28* (0.15)^2 * (1-0.15)^6$

$P(x = 2) = 0.2376$

So:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

$P(x \le 2) = 0.2725 + 0.3847 + 0.2376$

$P(x \le 2) = 0.8948$

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