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3 years ago

What is 8.05 × 10-5 in standard notation? 0.0000805 0.00000805 805,000 8,050,000

Mathematics

2 answers:

Tanya [424]3 years ago

4 0

Answer:

0.0000805

Step-by-step explanation:

$8.05*10^{-5}=\frac{8.05}{10^{5}}\\\\=\frac{8.05}{100,000}\\$

= 0.0000805

Katyanochek1 [597]3 years ago

3 0

Answer:

THe answer is A.

Step-by-step explanation:

I took the test :)

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Write an equation in standard form of the parabola that has the same shape as the graph of f(x)=5x^2or g(x)=-5x^2, but with a gi

pickupchik [31]

F(x)=5x^2 Has minimum (0,0)

g(x) = f(x) + 2 Shifts the graph two units up, then the minimum is 2+0=2.

h(x) = g(x+4) Shifts the graph four units left, then the minimum is at 0-4 = -4.

Then h(x) = 5(x+4)^2 + 2 has the minimum (-4,2)

And p(x) = -5(x+4)^2 + 2 has the maximum (-4,2)

4 0

3 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

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

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

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)

If solve n from equation (a) we got:

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

Part a

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 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.64$

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 $ME =\pm 0.02$ 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)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

Part b

For this case since we don't have a prior estimate we can use $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

8 0

3 years ago

Please help! will give brainliest !!

Veseljchak [2.6K]

Answer:

1, 2, 8, 128, 32768

Step-by-step explanation:

$recursive \: formula: \\ a _{n} = 2. {( a_{n - 1} )}^{2} \\ \\ a _{1} = 1...(given) \\ \\ a _{2} = 2. {( a_{2 - 1} )}^{2} = 2 {(a _{1} )}^{2} = 2 {(1)}^{2} \\ \\ a _{2} = 2\\ \\a _{3} = 2. {( a_{3 - 1} )}^{2} = 2 {(a _{2} )}^{2} = 2 {(2)}^{2} \\ \\ a _{3} = 8\\ \\a _{4} = 2. {( a_{4 - 1} )}^{2} = 2 {(a _{3} )}^{2} = 2 {(8)}^{2} \\ \\ a _{4} = 128\\ \\a _{5} = 2. {( a_{5 - 1} )}^{2} = 2 {(a _{4} )}^{2} = 2 {(128)}^{2} \\ \\ a _{5} = 32,768\\ \\$