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3 years ago

What is 2 gallons un quartz

Mathematics

2 answers:

photoshop1234 [79]3 years ago

6 0

2 gallons are 8 quarts since there are four to a gallon

kicyunya [14]3 years ago

4 0

Their are 4 quarts in a gallon so 2 x 4 = 8. Their are 8 quarts in 2 gallons. Hope that helps! :)

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I do have examples to help me but they make no sense :/

RideAnS [48]

First you would do 5^2 and that is 25. Next do 75+25 and that is 100. So I just did the top part next I would do the bottom. We know 5^2 is 25 and know we just divide 100 by 25 and that is 4. So your answer is 4

7 0

1 year ago

The mean age of swimers for all of these teams is 10. what does a large mad tell you.eade help

aev [14]

Answer: Larger the MAD tells us that the ages of swimmer are far from the mean age. Thus means is not a relevant indicator for the data.
Step-by-step explanation:
We know that the mean absolute deviation (MAD) helps to know whether the mean of a data is a worthy indicator for the data values .
The larger the MAD tells us the values are spread out far from the mean .
Also, larger the MAD makes the mean less worthy as an indicator of the data elements within the data set.

8 0

2 years ago

Apply the distributive property to factor out the greatest common factor. 35 + 14 =

Galina-37 [17]

Answer: approximately 49

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

From To Miles

balu736 [363]

Answer:

Good 4 U

Step-by-step explanation:

By Olivia Rodrigo

6 0

2 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago