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2 years ago

If the percent of discount is 75% and the sale price is $55.5 what is the original price?

Mathematics

1 answer:

gtnhenbr [62]2 years ago

8 0

Original price = (sale price*100):(100-percent of discount)

(55.5*100):(100-75)= 5550:25=222 $

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The table shows a pattern of exponents.

ahrayia [7]

Answer: c. divide the previous value by 5

Step-by-step explanation:

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2 years ago

What is the sum of the first five terms of a geometric series with a1 = 6 and r = 1/3?

tresset_1 [31]

The sum of the terms of a geometric sequence with common ratio lesser than 1 is calculated through the equation,

Sn = (a1) x (1 - r^n) / (1 - r)
Substituting the known values,
S5 = (6) x (1 - (1/3)^5) / (1 - 1/3) = 242/27
Thus, the sum of the first five terms is approximately equal to 8.96.

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3 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

2 years ago

Sorry I need help with these 3! Thank you to whoever takes the time to solve them!!!

mash [69]

1.)
Starts with 40:10, then cuz you have 80 buns, you need to x2, so it would be 20 cubes

2.)
I don’t know, sorry

3.)
4:2 and 60 in total
so 4+2 is 6 and 60/6 is 10
so it would be 40:20 so 20

hope this helps :)

8 0

3 years ago

1 HOUR TIME LIMIT

irina1246 [14]

Answer:

Your answer is 1500 foot pounds.

I hope this helps!

Step-by-step explanation:

Edg 2020

3 0

2 years ago

Read 2 more answers