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3 years ago

Math question no Guessing and Please show work

Mathematics

1 answer:

Nataly [62]3 years ago

6 0

What's the question?

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A person invest $8800 in an account at 7% interest compounded annually. When will the value of the investment be 13,200? (Type a

Nitella [24]

Answer:

21.43

Step-by-step explanation:

Turn 7 percent to a decimal: 0.07

8800 × 0.07 = 616

13,200÷616=21.4285 etc

Round two decimal places

21.43

7 0

2 years ago

Ow many significant figures are in the number 0.0040090?

Bumek [7]

5 significant figures but note that 0 is only non-significant when it begins a decimal its only significant when it's in between numbers or after a number

7 0

3 years ago

Help me ans should be well explained

Katen [24]

Finding diagonal sum

$\\ \sf\longmapsto 0.7+1.0+1.3$

$\\ \sf\longmapsto 0.7+2.3$

$\\ \sf\longmapsto 3.0$

Observing the square

x=0.9
y=0.6

$\\ \sf\longmapsto xy=0.9(0.6)$

$\\ \sf\longmapsto xy=0.54$

8 0

3 years ago

Read 2 more answers

Suppose your class sells gift wrap for $4 per package and greeting cards for $10 per package. Your class sells 205 packages in a

FromTheMoon [43]

Answer:

Step-by-step explanation:

1,084 divided by

3 0

3 years ago

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - α)% confidence interval for population mean (μ) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

σ = $60

MOE = $2

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago