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Furkat [3]
2 years ago
12

Roy's average balance checking account pays simple interest of 4.8% annually and he made $2.25 in interest last month. What was

Roy's average balance last month? A. 56.25 B. 5.63 C. 562.50 D. 5625.00
Mathematics
1 answer:
Leona [35]2 years ago
5 0

Answer:

It’s 5625.00 don’t listen to this other one

Step-by-step explanation:

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A 2-column table has 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is la
Anna [14]

Answer: The answer is f(x) = 0.8(2x).

i just did the quiz and got it right

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The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flas
Minchanka [31]
I have to think this again. be back later. X=2.22
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2 years ago
A farmer plants corn in 1/4 of his field. He plants white corn in 3/5 of the corn
bija089 [108]

Answer:

\frac{3}{20}

Step-by-step explanation:

Fraction of the total that is for corn (TC - Total corn):

TC=\frac{1}{4}

fraction of the corn section that is for white corn (WC - white corn in the corn seccion):

 WC=\frac{3}{5}

we need to find the fraction of the whole field that is for the white corn.

For this we need to find how much is \frac{3}{5} out of the \frac{1}{4} destinated to corn, and this will be the fraction of the total that is for white corn. We find this fraction by multiplying the fraction of corn (\frac{1}{4}) by the fraction of white corn in the corn section (\frac{3}{5}).

I will call the total fraction of white corn TWC, thus:

TWC=WC*TC=\\=\frac{3}{5} *\frac{1}{4} \\\\=\frac{3*1}{5*4} \\\\=\frac{3}{20}

the answer is: \frac{3}{20} of the whole field is planted with white corn

5 0
2 years ago
) The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students
katen-ka-za [31]

Answer:

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

Step-by-step explanation:

We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}). So we have T = 2.0930

The margin of error is:

M = T*s = 2.0930*0.16 = 0.3349.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851

The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

4 0
2 years ago
If 265 identical boxes, each containing 24 books weigh a total of 12,720 pounds, how much does each book weigh?
ivann1987 [24]
Each book ways 2 pounds


please like and rate
3 0
2 years ago
Read 2 more answers
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