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2 years ago

Which is the best example of a down payment?

Mathematics

1 answer:

Maru [420]2 years ago

8 0

Answer:

Step-by-step explanation:

Proof by some contradictions:

D is not a good example because she does not pay anything upfront

B is not a good example because he pays everything upfront with some of the money from a third party

A is not right because the 500 is a security deposit, which is a different concept from a down payment

Therefore the answer should be C, where money is payed upfront and payed after he receives the product.

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Solve the inequality. Check your solutions. 5t +6>21

Andreas93 [3]

Answer:

t > 3 or (3,∞)

Step-by-step explanation:

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2 years ago

What is the equation of the function that is graphed as line a?

Andrei [34K]

Answer:

Step-by-step explanation:

We see that the line intersects the y-axis at -1. As a result, we need to choose an equation with -1, either B or C. We see that the different is the slope. For choice B, the slope is $\frac{1}{3}$ , meaning that every unit up equates to 3 blocks to the right. Counting along the line, we see that this is correct.

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3 years ago

50 points What is the value of X in the triangle below? I need an explain and all work shown.

BARSIC [14]

The answer is C. Why? well, you should know SOH-CAH-TOA. IN THIS CASE YOU WILL USE SOH. SINE OVER HYPOTENUSE. THEN YOU SUBSTITUTE. SIN 45°= OPPOSITE/HYPOTENUSE. SIN45°=X/13 WHICH GIVES U CHOICE C.

3 0

3 years ago

Read 2 more answers

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

2 years ago

A certain population of mice is growing exponentially. After one month there are 48. After 2 months there are "192" mice. Write

REY [17]

Answer:

P(t) = 12e^1.3863k

Step-by-step explanation:

The general exponential equation is represented as;

P(t) = P0e^kt

P(t) is the population of the mice after t years

k is the constant

P0 is the initial population of the mice

t is the time in months

If after one month there are 48 population, then;

P(1) = P0e^k(1)

48 = P0e^k ...... 1

Also if after 2 months there are "192" mice, then;

192 = P0e^2k.... 2

Divide equation 2 by 1;

192/48 = P0e^2k/P0e^k

4 = e^2k-k

4 = e^k

Apply ln to both sides

ln4 = lne^k

k = ln4

k = 1.3863

Substitute e^k into equation 1 to get P0

From 1, 48 = P0e^k

48 = 4P0

P0 = 48/4

P0 = 12

Get the required equation by substituting k = 1.3863 and P0 = 12 into equation 1, we have;

P(t) = 12e^1.3863k

This gives the equation representing the scenario

8 0

2 years ago