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2 years ago

What is the quotient of 1+ 3 Over 4 and 2 over 5

Mathematics

1 answer:

dolphi86 [110]2 years ago

5 0

Answer:

2 1/2

Step-by-step explanation:

(These are fractions)

1+3 ÷ 2

4 5

(Simplify)

4 ÷ 2

4 5

(Same denominator)

20 ÷ 8

20 20

(Divide)

2 1/2

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If 430 pounds of pebbles cover 1500 square feet, how many pounds of pebbles are necessary to cover 75,000 square feet?

drek231 [11]

6500 pounds are necessary to cover 75000 square feet

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3 years ago

Can someone help plsss!

Roman55 [17]

Well the y-intercept is +2

And the slope of the blue line is 1/2

So y=1/2x+2

Slope is rise/run (up then ((right)))

And the intercept is found where the line intercects the y line.

7 0

2 years ago

Read 2 more answers

If one of the angles of a triangle is 130 degree then the angle between the bisectors of the other two angles can be

Viefleur [7K]

Answer: D) 155

Step-by-step explanation:

let <A, <B,<C are three angles in a triangle.

sum of the three angles in a triangle is 180degree.

<A+<B+<C=180

130+<B+<C=180 since one angle is given

<B+<C =180-130

<B+<C=50

(<B+<C)/2=25----(1)after bisecting sum of the angles

required angle =180-25=155

d is correct answer

8 0

3 years ago

Ms. Zab gave a group of 5 students a total of 18.75 liters of water for a science experiment. If each student in the group used

wlad13 [49]

Answer: The quantity of the water received by each student = 3.75 liters.

Step-by-step explanation:

Given: Total water = 18.75 liters

Total students =5

If each student in the group used the same amount of water for their experiment, then the quantity of the water received by each student = $\dfrac{\text{Total quantity}}{\text{Total students}}$

$=\dfrac{18.75}{5}\\\\=3.75\ liters$

The quantity of the water received by each student = 3.75 liters.

4 0

2 years ago

The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

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

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

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

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

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

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

5 0

2 years ago