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3 years ago

PLEASE HELP ———————-

Mathematics

1 answer:

Nuetrik [128]3 years ago

8 0

Answer:

It is a relationship of direct proportionality. When on a graph, traces a stright line. It will always produce a corresponding change.

e.x. y=2 and y=x It all corresponds into a stright line.

Hope it helps!

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How many four-digit numbers can be created using the digits 0-9? Note: A number cannot be repeated for different digits.

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6,561 is the right answer but good luck

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1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

uysha [10]

Answer:

1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$

Now we have everything in order to replace into formula (1):

$0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

Part 2

The original margin of error is given by:

$ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

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3 years ago

What function is increasing? will give brainlist !

Wewaii [24]

Answer:

Option B.

Step-by-step explanation:

Option A.

f(x) = $(0.5)^{x}$

Derivative of the given function,

f'(x) = $\frac{d}{dx}(0.5)^x$

= $(0.5)^x[\text{ln}(0.5)]$

= $-(0.693)(0.5)^{x}$

Since derivative of the function is negative, the given function is decreasing.

Option B. f(x) = $5^x$

f'(x) = $\frac{d}{dx}(5)^x$

= $(5)^x[\text{ln}(5)]$

= $1.609(5)^x$

Since derivative is positive, given function is increasing.

Option C. f(x) = $(\frac{1}{5})^x$

f'(x) = $\frac{d}{dx}(\frac{1}{5})^x$

= $\frac{d}{dx}(5)^{(-x)}$

= $-5^{-x}.\text{ln}(5)$

Since derivative is negative, given function is decreasing.

Option D. f(x) = $(\frac{1}{15})^x$

f'(x) = $-15^{-x}[\text{ln}(15)]$

= $-2.708(15)^{-x}$

Since derivative is negative, given function is decreasing.

Option (B) is the answer.

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Answer:

Step-by-step explanation:

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