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DanielleElmas [232]

4 years ago

5

Cassie is thinking of two numbers. Adding 4 times the first number and 6 times the second number gives a total of 28. Also, addi

ng 5 times the first number and 6 times the second number gives 29. What are the two numbers?

1 answer:

aleksandrvk [35]4 years ago

3 0

If you try 6x4, you get 24. When you add 4, it = 28. The same thing with 6x4=24+5=29

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Which of these is NOT an integer

Tema [17]

Hi there!

»»————-　★　————-««

I believe your answer is:

Option A

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Here’s why:

⸻⸻⸻⸻

$\boxed{\text{Option A:}}\\\\3^{-5}\\\\\rightarrow \frac{1}{3^5}\\\\\rightarrow \frac{1}{243}\\\\\ \text{This is \textbf{NOT} an integer.}}$

⸻⸻⸻⸻

$\boxed{\text{Option B:}}\\\\-3^5\\\\\rightarrow -3 * -3 * -3 * -3 * -3 \\\\\rightarrow \boxed{-243}\\\\\\\text{This \textbf{IS} an integer.}$

⸻⸻⸻⸻

$\boxed{\text{Option C:}}\\\\\frac{1}{3}^{-5} \\\\\rightarrow \frac{1}{(\frac{1}{3})^5}\\\\\rightarrow \frac{1}{\frac{1}{243} }\\\\\rightarrow \boxed{243}\\\\\\\text{This \textbf{IS} an integer.}$

⸻⸻⸻⸻

$\text{Option \textbf{A} does not simplify into an integer.}$

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»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

7 0

3 years ago

When factored completely the expression p^4-81 is equivalent to

spayn [35]

$\displaystyle\\
p^4 - 81 =\\\\
 =p^4-3^4 =\\\\
= \Big(p^2\Big)^2 - \Big(3^2\Big)^2 =\\\\
=\Big(p^2+3^2\Big)\Big(p^2-3^2\Big)=\\\\
=\boxed{\bf \Big(p^2+3^2\Big)\Big(p+3\Big)\Big(p-3\Big)}$

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What is 35% of 60? just had to be 20 words

Vladimir [108]

Answer:

35% of 60 = 21

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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 $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

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

Step-by-step explanation:

<h3>(a)</h3>

$P(A_1 \; \cap \; A_2) = 0$ 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.

<h3>(b)</h3>

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:

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

<h3>(c)</h3>

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:

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

<h3>(d)</h3>

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:

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

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Brrunno [24]

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