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4 years ago

Find two-three digit numbers whose sum is a multiple of 504 and the quotient is a multiple of 6

Mathematics

1 answer:

fiasKO [112]4 years ago

4 0

79+5=84
84*6= 504
So the numbers should be 79 and 5
If that's what they r asking for and that's what I believe they r

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A car salesman had $65,100 in sales. He earned $1,953 in commission. What percent commission did he earn. Explain your answer.

ahrayia [7]

Answer:

1. The percent commission earned is 3%.

2. The loan period is 3.29 years.

Step-by-step explanation:

1. Salesman has $65,100 in sales. He earned $1,953 in commission.

Let the percent commission earn be x%

Therefore, x% of the sales equals $1,953

$\frac{x}{100} of $65,100 = $1,953$

$\frac{x}{100} * 65100 = 1953\\\frac{65100x}{100} = 1953$

We cross multiply

$\frac{65100x}{100} = 1953\\65100x = 1953 * 100\\65100x = 195300$

Divide both side by the coefficient of 'x' (65100)

$\frac{65100x}{65100} = \frac{195300}{65100} \\x = 3$

Therefore, the percent commission earned is 3%

2. Interest (I) = $299 Principal (P) = $1300 Rate (R) = 7%

The formula for finding interest is given as: $I = \frac{PRT}{100}$

Therefore, substituting into the formula, we have:

$299 = \frac{1300 * 7 * T}{100}$

We are finding the time it takes the loan to earn an interest of $299

$299 = \frac{1300 * 7 * T}{100} \\299 = \frac{9100T}{100}$

We cross-multiply:

$299 * 100 = 9100T\\29900 = 9100T$

Divide both side by the coefficient of T (9100)

$\frac{29900}{9100} = \frac{9100T}{9100}\\T = 3.29$

Therefore, the time taken for the loan to earn such interest is approximately 3.29 years

4 0

4 years ago

Which value cannot represent the probability of an event occurring? StartFraction 1 over 100 EndFraction 0. 29 85% Three-half.

Serjik [45]

Every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.

Given to us,

a.) $\dfrac{1}{100}$

b.) 0.29

c.) 85%

d.) $\dfrac{3}{2}$

The probability help us to know about the probability of specific events occurring.

For a sure event, the probability is always 1,

while for an event that will never happen the probability is always 0.

Thus, probability(p), $\bold{1 \geq p\geq 0}$ .

Now looking at the options,

a.) $\dfrac{1}{100}$ = 0.01

b.) 0.29

c.) 85% = 0.85

d.) $\dfrac{3}{2}$ = 1.5

Now comparing each option with $\bold{1 \geq p\geq 0}$ .

Therefore, the only option which is not feasible is option d that is 1.5.

Hence, every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.

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brainly.com/question/795909

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