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3 years ago

Circle any 2 factors of 400

Mathematics

2 answers:

Pie3 years ago

7 0

1,2,4,5,8,10 that should be enough

Alexxandr [17]3 years ago

6 0

A factor is a group of 2 or more numbers that can be multiplied to make a greater number.

The factors for 400 include:

1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200

When paired together:

(1, 400), (2, 200), (4, 100), (5, 80), (8, 50), (10,40), (16, 25), (20,20)

Prime factors:

1, 2, 5

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As – 16a,<br> as – 4a3,<br> 2a3 + 8a - 4a2 - 16

vfiekz [6]

Answer:

1) as-16a

taknig a as common

=a(s-16)

2)as-4a³

taking a as common

=a(s-4a²)

3)2a³+8a-4a²-16

taking common terms

=2a(a²+4)-4(a²+4)

=(2a-4)(a²+4)

again taking common

=2(a-2)(a²+4)

i hope this will help you :)

7 0

3 years ago

Andy and Sam are saving money to go on their senior trip. The amount of money that Andy will have at the end of each week, w, ca

Zanzabum

I think the answer is (400w)

6 0

3 years ago

Read 2 more answers

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

3 years ago

The cost of a baguette and a cup is 1.45 the cost of 2 baguettes and a cup is 2.10 what is the cost of the cup

olga2289 [7]

Answer:

0.80

Step-by-step explanation:

2.10-1.45=0.65

0.65=1 baugette

1.45-0.65=0.8

0.8=1 cup

5 0

4 years ago

Ana lee has 30 gum balls. If one sixth are green, and one third are red,are there

Andrew [12]

Red I know because the smaller the number the bigger the fraction

4 0

3 years ago