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Ray Of Light [21]
3 years ago
10

Solving Linear equations 25 =x + 19

Mathematics
2 answers:
Pie3 years ago
6 0
25=x + 19
-19 -19


6=x
soldi70 [24.7K]3 years ago
3 0

Steps to solve:

25 = x + 19

~Subtract 19 to both sides

6 = x

Best of Luck!

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suppose that a company claims that its boxes of breakfast cereal contain 15 ounces of cereal on average, with a standard deviati
olga nikolaevna [1]

Answer: 15 ounces


Step-by-step explanation:

It is given that a company claims the average cereal in boxes of breakfast cereal = 15 ounces

And we know that the mean amount is nothing but the average amount of any data.

Mean is synonym word of average .

The statistical mean is the mean or average that is used to find the central tendency of the data in any question.

Thus, the expected mean amount of cereal per box = 15 ounces.

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Two angles are complementary. The larger
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Answer:

Step-by-step explanation:

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3 years ago
Helpp!!!i am very confused with this unit
ohaa [14]
24*3.14= 75.36= 15.37 in^2 is your answer

help with rounding numbers:
four or less, let it rest ( 4 and less let the number be the same and change the numbers after it into zeros)
five or more, change the score. ( five and more change the number one up and change the numbers after it into zeros)
6 0
3 years ago
Sali throws an ordinary fair 6 sided dice once.
xenn [34]

a) 1/6

b) 1/36

c)

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

Step-by-step explanation:

a)

The probability of a certain event A to occur is given by

p(A)=\frac{a}{n}

where

a is the number of successfull outcomes, in which event A occurs

n is the total number of possible outcomes

In this problem, the event is

"getting a 6 when throwing a dice once"

We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have

n=6

The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore

a=1

So, the probability of this event is

p(6)=\frac{1}{6}

b)

In this case instead, we are throwing the dice twice.

The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,

p(AB)=p(A)\cdot p(B)

where

p(A) is the probability that event A occurs

p(B) is the probability that event B occurs

Here we have:

- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is

p(A)=\frac{1}{6}

- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so

p(B)=\frac{1}{6}

Therefore, the probability of getting a 6 on both throws is:

p(66)=p(6)\cdot p(6)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}

c)

In this problem, we have:

- A dice that is thrown once

- A coin that is also thrown once

The dice has 6 possible outcomes, as we stated in part a):

1, 2, 3, 4, 5, 6

While the coin has two possible outcomes:

H = head

T = tail

So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.

Doing so, using the following notation:

1H (getting 1 with the dice, and head with the coin)

The possible outcomes are:

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

So, we have a total of 12 possible outcomes.

4 0
3 years ago
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