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3 years ago

lisa says that she made twice as many successful shots as sue but scored the same number of points how is this possible

Mathematics

1 answer:

sweet [91]3 years ago

4 0

Lisa could have made 2-pointers while Sue made 3-pointers.

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Lacey baked 20 cookies with 2 scoops of flour. With 5 scoops of flour, how many cookies can Lacey bake? Assume the relationship

Komok [63]

Answer:

The number of cookies baked is 50

Step-by-step explanation:

Given as:

Lacey baked 20 cookies with 2 scoops of flour

∵ For 2 scoops of flour , The number of cookies baked is 20

so,For 1 scoops of flour , The number of cookies baked is $\frac{20}{2}$

∴ For 5 scoops of flour , The number of cookies baked is $\frac{20}{2}$ × 5

I.e Number of baked cookies = 10 × 5 = 50

Hence The number of cookies baked is 50 Answer

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3 years ago

Trevor and Kevin bought baseball tickets with 2/3 of their money. They spent 1/4 of the remainder on popcorn and peanuts for the

VladimirAG [237]

70*.66=46.2

46*0.25=11.55

11.55+46.2= 57.75

70-57.75=12.25

12.25 dollars remaining from after the game

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3 years ago

Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three f

GREYUIT [131]

Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

He has a 3/4 probability of success.

This means that $p = \frac{3}{4} = 0.75$

Giannis shoots three free throws

This means that $n = 3$

What is the probability that he succeeds on at least two of them

$P(X \geq 2) = P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219$

$P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438$

84.38% probability that he succeeds on at least two of them

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3 years ago

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Lana71 [14]

Answer:x^6/4

Step-by-step explanation:

Simplifying the steps

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3 years ago