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3 years ago

Help please I got to pass

Mathematics

1 answer:

mixer [17]3 years ago

4 0

Answer:

A'(5,3)

Step-by-step explanation:

First, you must understand that A(5,3) is the pre-image and that A' is what we are looking for which is the image.

With that in mind, you translate the preimage by adding or subtracting from the x and y values.

(x+4, y-3)

To find the x value of the pre-image, you will add 4 to the preimages' x value

Pre-image: A(1,6)

To find the y value, you subtract the preimages' y value by 3.

Hope this helps!

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Identify the standard form of a circle equation x^2+2x+y^2-2y=2

Bas_tet [7]

Answer:

The standard from of the expression is: $(x +1)^2 + (y-1)^2 = (2)^2$

Step-by-step explanation:

Here the given expression is :

$x^2+2x+y^2-2y=2$

Now, the standard form of a circle is given as :

$(x-h)^2 + (y -k)^2 = r^2$

Here, (h,k) = Coordinates of Center, r = Radius

Also, use the algebraic identity:

$(a \pm b)^2 = a^2 + b^2 \pm 2ab\\$

Now, converting the given expression in the standard form, we get:

$x^2+2x+y^2-2y=2$

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$x^2+2x+y^2-2y +2=2 + 2\\\implies x^2+2x + 1 + y^2-2y + 1 = 4\\\implies (x^2+2x + 1 )+ (y^2-2y + 1) = 4\\\implies (x +1)^2 + (y-1)^2 = (2)^2$

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3 years ago

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2005 was $110,000. A sample of 81

Sholpan [36]

Answer:

a) We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:

Null hypothesis: $\mu \leq 110000$

Alternative hypothesis: $\mu > 110000$

b) $z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$

c) $p_v =P(Z>2.5)=0.0062$

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.

Step-by-step explanation:

Information given

$\bar X=12000$ represent the sample mean for the yearly income in 2006 for the dentists

$\sigma= 36000$ represent the sample population deviation

$n=81$ sample size

$\mu_o =110000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value for the test

a) Hypothesis to verify

We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:

Null hypothesis: $\mu \leq 110000$

Alternative hypothesis: $\mu > 110000$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

b) Statistic

When we replace the data given we got:

$z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$

c) P value

Now we can find the p value using the fact that we are conducting a right tailed test:

$p_v =P(Z>2.5)=0.0062$

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.

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