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3 years ago

Many 4 digit numbers can be formed using only the numbers 3 2 7 6 0 4 9 if

1 answer:

6 0

Hi!

The combination of 4 digit numbers using those numbers are: 3, 276; 3, 762; 3, 267; 3, 726; 3, 672 etc. This is pretty easy, now you try.

Hope I helped! Have a nice day!

You might be interested in

What is the balance after 15 years in a savings account that earns 2% interest compounded bimonthly when the initial deposit is

Ierofanga [76]

Answer:

$1348.07

Step-by-step explanation:

Hello!

<h3>Compound Interest Formula: $A = P(1 + \frac rn)^{nt}$ </h3>

A = Account Balance
P = Principle/Initial Amount
r = Rate of Interest (decimal)
n = Number of times compounded (per year)
t = Number of Years

<h3>Given Information</h3>

Account Balance = ?
Principle Amount = $1000
Rate of Interest = 0.02

Why is the Rate 0.02?

This is because we are gaining money, so the multiplier should be greater than 1. We already added 1, which is 100% so you simply add the 0.02 for the extra 2%.

Number of times compounded per year = 6

This is because it is being compounded bi-monthly, or once every 2 months. 12 months divided by 2 months is 6 months, so 6 times a year.

Number of years = 15

<h2>Solve </h2>

Solve by plugging in the given values into the formula.

$A = P(1 + \frac rn)^{nt}$
$A = 1000(1 + \frac {0.02}{6})^{6*15}$
$A = 1000(1 + 0.00333...)^{90}$
$A = 1000(1.00333...)^{90}$
$A = 1000(1.145743)$
$A \approx 1457.43$

This is really close to the first option, and since there is rounding involved with the repeating decimal, the first option should be correct.

The answer is $1348.07.

3 0

2 years ago

What is m∠1?............

gogolik [260]

the answer is 148

hope that helped lol

8 0

3 years ago

Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size

julsineya [31]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641$

And rounded up we have that n=16641

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

For this case we can use as estimator for the proportion $\hat p =0.5$ , since we don't have any other previous info. And replacing into equation (b) the values from part a we got: