Ask question

Ask question

All categories

3 years ago

11

Pete's checking account showed a balance at the beginning of the month of $180. After 3 days the checking account showed a balan

ce of $160. Write a linear equations to represent the total in the checking account f(x) according to the day x.

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

180 - 172 =$8 8/3=2.56

You might be interested in

Which is the difference of 21 1/4 - 18 2/4? A:2 1/4. B:2 2/4. C:2 3/4. D:3 1/4.

seropon [69]

Answer:

2 3/4 C

Step-by-step explanation:

$(21+\frac{1}{4})-(18+\frac{2}{4})\\=21-18+\frac{1}{4} -\frac{2}{4} \\=3-\frac{1}{4} \\=\frac{12-1}{4} \\=\frac{11}{4} \\=2\frac{3}{4} \\or\\21\frac{1}{4} -18\frac{2}{4} \\=\frac{85}{4} -\frac{74}{4} \\=\frac{85-74}{4} \\=\frac{11}{4} \\=2\frac{3}{4}$

3 0

3 years ago

Can someone please help me get 2 brainliest?

Rasek [7]

Answer:

what?

Step-by-step explanation:

8 0

3 years ago

What is the final amount if 784 is decreased by 1% followed by a 4% increase?

Ilya [14]

Answer:

3136

Step-by-step explanation:

Thats the answer please I don't have time to write the explanation

5 0

3 years ago

What is the greatest common factor of 52x5 and 12x4

Pani-rosa [81]

"A. 4x^4 is the answer so try that."

7 0

3 years ago

Read 2 more answers

In a random sample of 8 people, the mean commute time to work was 33.5 minutes and the standard deviation was 7.2 minutes. A 95

Elina [12.6K]

Answer: Margin of error = 6.58

Confidence interval = (26.91, 40.08)

Step-by-step explanation:

Since we have given that

Sample size n = 8

Sample mean = 33.5 minutes

Population standard deviation = 9.5 minutes

At 95% confidence interval,

α = 0.05

t = 1.96

So, Margin of error is given by

$t\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{9.5}{\sqrt{8}}\\\\=6.58$

Confidence interval would be

Lower limit:

$\bar{x}-6.58\\\\=33.5-6.58\\\\=26.91$

Upper limit:

$\bar{x}+6.58\\\\=33.5+6.58\\\\=40.08$

Hence, the interval would be (26.91, 40.08)

But at standard deviation 7.2 minutes, the confidence interval was (27.5, 39.5)

Confidence interval using the standard normal distribution is wider than the confidence interval using t distribution.

3 0

3 years ago