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3 years ago

What decimal is equivalent to negative 3/8

Mathematics

2 answers:

hoa [83]3 years ago

8 0

Answer:

-0.375

Step-by-step explanation:

If you have a scientific calculator just plug in $-\frac{3}{8}$ then switch it to decimal form

$-\frac{3}{8} = -0.375$

If you need help going from fraction to decimal form all you need to do is divide -3 by 8 = -0.375

Fittoniya [83]3 years ago

7 0

Answer:

-6/16

Step-by-step explanation:

Step 1: Find equivalent to -3/8

(-3*2) / (8*2)

-6/16

Answer: -6/16

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A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each

klasskru [66]

Answer:

$P(X_s^c|X_F) =0.2$

$P(X_s^c|X_F) =0.31$

$P(X_s^c|X_F) =0.331$

Step-by-step explanation:

From the given information:

Let represent $X_F$ as the first player getting an ace

Let $X_S$ to be the second player getting an ace and

$\sim X_S$ as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;

$P(\sim X_S| X_F) = 1 - P(X_S|X_F)$

$P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}$

Let's determine the probability of getting an ace in the first player

i.e

$P(X_F) = 1 - P(X_F^c)$

$= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}$

$= 1 - \dfrac{n-1}{2(2n-1)}$

$= \dfrac{3n-1}{4n-2} --- (1)$

To determine the probability of the second player getting an ace and the first player getting an ace.

$P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}$ $P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}$

$P(X_sX_F) = \dfrac{n}{2n -1}---(2)$

$P(X_s|X_F) = \dfrac{2}{1}$

$P(X_s|X_F) = \dfrac{2n}{3n -1}$

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

$P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}$

$P(X_s^c|X_F) = \dfrac{n-1}{3n -1}$

Therefore;

for n= 2

$P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}$

$P(X_s^c|X_F) = \dfrac{1}{6 -1}$

$P(X_s^c|X_F) = \dfrac{1}{5}$

$P(X_s^c|X_F) =0.2$

for n= 10

$P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}$

$P(X_s^c|X_F) = \dfrac{9}{30 -1}$

$P(X_s^c|X_F) = \dfrac{9}{29}$

$P(X_s^c|X_F) =0.31$

for n = 100

$P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}$

$P(X_s^c|X_F) = \dfrac{99}{300 -1}$

$P(X_s^c|X_F) = \dfrac{99}{299}$

$P(X_s^c|X_F) =0.331$

8 0

3 years ago

5.) In a basketball game 96 points were scored. The winners scored twice as many as the

labwork [276]

Answer:

losers= 32 points winners=64

Step-by-step explanation:

winners:2x

losers:x

2x+x=3x

96/3=32

losers=32*1

Winners=32*2

7 0

4 years ago

The probability that a company will launch the product A and B are 0.45 and 0.60 respectively, in main while, probability that b

Brrunno [24]

Answer:

a) what is the probability that Neither will of these products launch ?

= 0.30

b) At least one product will be launched ?

= 0.70

Step-by-step explanation:

From the above question, we have the following information:

P(A) = 0.45

P(B) = 0.60

P(A ∩ B) = P(A and B) launching = 0.35

Step 1

We find the Probability that A or B will launch

P (A ∪ B) = P(A) + P(B) - P(A ∩ B)

= 0.60 + 0.45 - 0.35

= 1.05 - 0.35

= 0.70

a) what is the probability that Neither will of these products launch ?

1 - Probability ( A or B will launch)

= 1 - 0.70

= 0.30

b)At least one product will be launched?

This is equivalent to the probability that A or B will be launched

P (A ∪ B) = P(A) + P(B) - P(A ∩ B)

= 0.60 + 0.45 - 0.35

= 1.05 - 0.35

= 0.70

3 0

3 years ago

Order of operations 11 + 14 ÷ 2 - 1

grandymaker [24]

Answer:

Step-by-step explanation:

11+14÷2-1

11+7-1

18-1

8 0

3 years ago

Read 2 more answers

the temperature was 8 degrees Fahrenheit below zero in Rose to 10 degrees Fahrenheit what is the temperature now

Svetach [21]

Answer: the answer is 2ºf

Step-by-step explanation:

-8ºf+10ºf = 2ºf

5 0

3 years ago