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Dovator [93]
3 years ago
12

It rained 0.72 inches on Monday. On Tuesday, it rained 0.34 inches less than on Monday.

Mathematics
2 answers:
vodomira [7]3 years ago
8 0

Answer:0.38 inch

Step-by-step explanation: If it rains 0.72 inch on Monday and on Tuesday it rains 0.34 less inch then, you would have to subtract them to get 0.38 inch aka answer.

Zigmanuir [339]3 years ago
4 0

Answer: The answer is 0.38 inches

Step-by-step explanation:

The question said that on Tuesday, it rain 0.34 less than on Monday which mean to subtract. Therefore, 0.72-0.34=0.38 inches.

I hope it helps :)

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Which equation can be used to solve X in the following diagram?
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A) 6x+12x+90=180

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18x=90

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x=5

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3 years ago
Write a compound interest function to model each situation. Then find the balance after the given number of years.
andriy [413]

Answer:

See explanation

Step-by-step explanation:

The standard compound interest formula is A = P(1+\frac{r}{n})^{nt} where:

P is the principal amount

r is the interest rate (typically as a percentage)

t is the time

n is the times compounded per unit of time

So,

1)  A = 50000(1+\frac{0.03}{12})^{(12)(6)} =59847.42

2) A = 43000(1+\frac{0.05}{1})^{(1)(3)} =49777.88

3) A = 65000(1+\frac{0.06}{4})^{(4)(12)} =132826.08

You should check my answers though, I may have mixed up some terms.

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3 years ago
A smartphone costs ​$120. Sales tax is 55​%. What is the total​ cost, including​ tax?
hjlf
You multiply  $120 by .55 to get the cost for tax which is $66 you add that to $120 to get $186
6 0
3 years ago
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Which fraction is less then 1/3 or 1/5
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1/12

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Which expression should you simplify to find the 90% confidence interval,
Alex787 [66]

Answer:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287

0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

Step-by-step explanation:

The estimated proportion of interest is \hat p=0.45

We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value is:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the true population proportion is interest is given by this formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing the values provided we got:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287

0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

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