-7+10x=54 9y-5x=81 Solve please

The inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y

Cerrena [4.2K]

We have been given that the inverse variation $y=\frac{3\times 10^8}{x}$ shows the relationship between wavelength in meters, x, and frequency, y. We are asked to find the wavelength for radio waves with frequency $3\times 10^9$ .

To find the required wavelength, we substitute $y=3\times 10^9$ in our given equation and solve for x as:

$3\times 10^9=\frac{3\times 10^8}{x}$

$x=\frac{3\times 10^8}{3\times 10^9}$

We can see that both numerator and denominator has 3, so we can cancel it out.

$x=\frac{1\times 10^8}{1\times 10^9}$

Using quotient rule of exponents $\frac{a^m}{a^n}=a^{m-n}$ , we will get:

$x=1\times 10^{8-9}$

$x=1\times 10^{-1}$

Therefore, the wavelength for radio waves would be $1\times 10^{-1}$ meters and option B is the correct choice.

3 0

3 years ago

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If u have 25% how many squares would you shade if there were 6 in total?

egoroff_w [7]

Answer:

1.5 squares

Step-by-step explanation:

4 0

3 years ago

What's the q1 and q2????

Evgen [1.6K]

Answer:

Q1-298

Q2-422

Step-by-step explanation:

8 0

2 years ago

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Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

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

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago

Is 9-z polynomial ?

storchak [24]

Answer:

No

Step-by-step explanation:

5 0

3 years ago