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3 years ago

2/3 - 2/9 what is the answer ???

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

8 0

Answer:

0.44

Step-by-step explanation:

2/3 - 2/9 = 0.44

neonofarm [45]3 years ago

7 0

Answer: 4/9

Step-by-step explanation:

Covert them into having the same denominator so 2/3 into 6/9 then subtract 6/9 -2/9. 6-2 is 4 and you keep the denominator the same which makes the answer 4/9.

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Using the formula in model 1, choose the correct answers for the total amount and amount of interest earned in the following com

DIA [1.3K]

The total amount is $ 1015.82 and interest amount is $ 165.82

Solution:

The formula for amount when interest is compounded annually is:

$\mathrm{A}=P\left(1+r}\right)^{n}$

Where,

"A" is the total amount

"P" is the principal

"r" is the rate of interest in decimal form

"n" is the number of years

From given, $850 at 2% for 9 years, compounded annually

P = 850

t = 9 years

$r = 2 \% = \frac{2}{100} = 0.02$

Substituting the given values we get,

$A = 850(1+0.02)^9\\\\A = 850 \times 1.02^9\\\\A = 850 \times 1.1950\\\\A = 1015.82$

Thus total amount is $ 1015.82

Interest amount = Total amount - principal

Interest amount = 1015.82 - 850

Interest amount = 165.82

Thus total amount earned is $ 1015.82 and interest amount is $ 165.82

7 0

3 years ago

2. How these concepts are used indifferent situations?

Neko [114]

If you tell me the concepts you’re referring to I might be able to help

8 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of