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3 years ago

if a player has the cards 1, -7, and 4, find two ways to get five. explain and give the equation (absolute value can be used)

Mathematics

1 answer:

dimulka [17.4K]3 years ago

3 0

-7+4+1 Hope u appreciate my answer

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On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1

muminat

Answer:

$Probability = \frac{1}{24}$

Step-by-step explanation:

Given

$Oranges = 3$

$Apples = 5$

$Banana = 2$

$Total = 10$

Required

Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

$Probability = P(Orange) * P(Apple) * P(Banana)$

$Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}$

The difference in the numerator is as a result of picking the fruit without replacement

$Probability = \frac{30}{720}$

$Probability = \frac{1}{24}$

6 0

2 years ago

Indica el valor del 3 en cada uno de los siguientes números 35.097 _ 753.978 _ 7.543 _ 312.501

lara31 [8.8K]

Free points pls and thank you

5 0

3 years ago

Hot Shot Electronics is designing a packing box for its new line of Acoustical Odyssey speakers. The box is a rectangular prism

Whitepunk [10]

V= W × H × L
81000 = 30 × H × 45
81000 = 1350× H
81000/1350 = 1350/1350× H
H = 60 cm

8 0

3 years ago

Select the equation that represents the problem let x represent the unknown number

djyliett [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable X is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, p = 0.09.

An item is defective irrespective of the others.

The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of X is 0.36.

Compute the variance of X as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of X is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

4 0

2 years ago