All categories

3 years ago

The measures of Zx and Zy are equal. Which is the value of m?

Mathematics

1 answer:

Ede4ka [16]3 years ago

3 0

Z because y=Mx+b so whatever is before x is the answer in this case it’s z

You might be interested in

Which expression is equivalent to 5(c + 2(c+d)

Fofino [41]

15c+10d that’s the answer

5 0

3 years ago

On a recent day, 8 euros were worth $11 and 32 euros were worth $44. Enter an equation of the form y = kx to show the relationsh

navik [9.2K]

Answer:

On a recent day, 8 euros were worth $11 and 40 euros were worth $55. Enter an equation of the form y = kx to show the relationship between the number of euros and the value in dollars. First solve k which is the slope so k = ( 55- 11) / ( 40 -8) = 1.375

Y= 1.375x

Step-by-step explanation:

4 0

3 years ago

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

3 years ago

Simplify: (6 + 2iX6 - 21)

snow_tiger [21]

If simplify it would be -15 + 12i

7 0

3 years ago

Read 2 more answers

Is 660 ten times as many as 600?

irina1246 [14]

Im pretty sure (like 99.9999999% sure) that 660 is not 10 times as much as 600 because when you multiply 600 by 10 you don't get 660. You get 6000.

3 0

3 years ago

Read 2 more answers