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3 years ago

Plz help!!!!!!!!! Due soon!

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

520 Students

Step-by-step explanation:

If you multipy 650 by .8 you will get 520 and that is the percentage of students that is going.

alekssr [168]3 years ago

5 0

Answer:

520 students will attend the field trip.

Step-by-step explanation:

The percentage of a number can be found by multiplying the percentage by the amount.

In this case, 650 times 80% is 520.

We can check our answer by dividing 520 by 650 and converting the decimal into a percentage.

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Crystal earns $4.00 per hour mowing lawns. how much does crystal earn if she works 2 hours and 30 minutes?

zlopas [31]

10 dollars i got this by finding 4 times 2 is 8 and half of 4 is 2 so it is exactly 10 dollars

7 0

3 years ago

Which expressions show the expanded form for 200.06 choose all answers that are correct.

Anit [1.1K]

a. 2 x 100 0 x 1 0 x 1/10 6 1/100b. 2 x 100 0 x 1 0 x 1/100 6 1/1000

Hope it helps! :D

Btw, where is B?

4 0

3 years ago

Between which two consecutive integers does ???147 lie?

BartSMP [9]

If those question marks are supposed to be a radical, then 12 and 13. The square root of 147= 12.12, which puts it between 12 and 13.

3 0

3 years ago

Let f(x) = 12 over the quantity of 4 x + 2 . Find f(−1). (1 point) a −6b −2c 2d 6

antoniya [11.8K]

Answer:

f(-1) = -6

Step-by-step explanation:

f(x) = 12 / ( 4x+2)

Let x = -1

f(-1) = 12 / ( 4*-1+2)

= 12 / (-4+2)

= 12 / -2

= -6

5 0

3 years ago

Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int

meriva

Answer:

b) 95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are $(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })$ the 95 percent confidence interval for the population mean 'μ'

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

$(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })$

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

Conclusion:-

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

6 0

3 years ago