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svp [43]
3 years ago
14

Shama’s age is three times the age of her son Ankit. 10 years ago Shama

Mathematics
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

Step-by-step explanation:

Let sharma age be S and Ankit age be A

S = 3a

(s-10) = 5(a-10)

substitute 1 in 2

3a-10 = 5a -50

40 = 2a

a = 20

s = 3 * 20 = 60

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Beth filled 72 jars with paint. If each jar holds 1 pint of paint, how many gallons of paint did beth use?
Paul [167]

Answer:

9 gallons were used

Step-by-step explanation:

There are 8 pints to 1 gallon

72 jars holding 1 pint each means 72 pints total

72/8=9

Beth used 9 gallons

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5 0
3 years ago
Read 2 more answers
If the sales tax rate is $8.25 per $100.00.What is the sales tax if the sale amount is $54.80?
Brums [2.3K]

Answer: $4.52

Step-by-step explanation:

From the question, sales tax rate is $8.25 per $100, this means that for every $100 item sold, $8.25 will be paid as sales tax.

Sales tax for 54.80 = x

Step 2: write the equation as;

8.25 = $100

Sales tax for $1 item will be (divide both sides by 100

$8. 25/ 100 = $0.0825

Step 3: determine the sales tax on $1 item, that is if sales tax $1 item = $0.0825

Sales tax for $54.80 item will be 0.0825 x 54.80

= $4.52

7 0
3 years ago
The area of a rectangular window is 7426 cm
soldi70 [24.7K]

hi <3

area = length x width

substitute the values in:

7426 = length x 79

rearrange to get the answer:

length = 94 cm

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7 0
2 years ago
Solve 3sinx = 2cos^2x .
goldenfox [79]
3 sin x = 2 ( 1 - sin² x )
3 sin x = 2 - 2 sin² x
2 sin² x + 3 sin x - 2 = 0
Substitution: t = sin x
2 t² + 3 t - 2 = 0
t 1/2 = \frac{-3+/- \sqrt{9+16} }{4} =  \frac{-3+/-5}{4}
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
Answer:
x 1 = π / 6 + 2 kπ,       x 2 = 5 π / 6 + 2 k π,   k ∈ Z 
5 0
3 years ago
The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d
Colt1911 [192]

Answer: 34%

Step-by-step explanation:

According to the Empirical rule,

About 68% of the population lies with in one standard deviation from the mean.

i.e. About  34% of the population lies above one standard deviation from the mean .

and About 34% of the population lies below one standard deviation from the mean.

Given : The distribution of the number of daily requests is bell-shaped ( i.e. Normally distribution) and has a mean of 60 and a standard deviation of 11.

i.e. \mu=60\ \ \sigma=11

Using the Empirical Rule rule,  34% of the population of lightbulb replacement requests lies above one standard deviation from the mean .

i.e. About 34% of the population of lightbulb replacement requests lies between \mu and \mu+\sigma

i.e. About   34% of the population of lightbulb replacement requests lies between 60 and 60+11

i.e. About  34% of the population of lightbulb replacement requests lies between 60 and 71

Hence, the approximate percentage of lightbulb replacement requests numbering between 60 and 71 = 34%

4 0
3 years ago
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