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Leno4ka [110]
3 years ago
9

The ratio of the number of apples to the number of pears in a basket was 3:5. After 7more apples were added to the basket and 9

pears were removed ,there were an equal number of apples and pears .how many apples and how many pears were originally in the basket
Mathematics
1 answer:
atroni [7]3 years ago
5 0

Answer:

Apples = 24

Pears = 40

Step-by-step explanation:

Let u be the unit of fruit in the basket,

number of apples: 3u +7

number of pears: 5u -9

now to find u, we equate the two equations.

3u+7=5u-9

Now simplify.

5u-3u=7+9

2u=16

u=8

Now we can find the original number of apples and pears by substituting u.

so original number of apples = 3u = 3(8) = 24

and original number of pears = 5u = 5(8) = 40

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Answer:

x=3

Step-by-step explanation:

When you substitute x = 3 to both equations, you get y = -4 for both so that is a solution

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3 years ago
If discriminant (b^2 -4ac>0) how many real solutions
MaRussiya [10]

Answer:

If Discriminant,b^{2} -4ac >0

Then it has Two Real Solutions.

Step-by-step explanation:

To Find:

If discriminant (b^2 -4ac>0) how many real solutions

Solution:

Consider a Quadratic Equation in General Form as

ax^{2} +bx+c=0

then,

b^{2} -4ac is called as Discriminant.

So,

If Discriminant,b^{2} -4ac >0

Then it has Two Real Solutions.

If Discriminant,b^{2} -4ac < 0

Then it has Two Imaginary Solutions.

If Discriminant,b^{2} -4ac=0

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2 years ago
A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

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Expected reward for this approach

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Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

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<span>2:3:4
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2x = 12
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Therefore, the length of the longest side would be 4x = 4(6) = 24 cm, third option.</span>
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