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4 years ago

Can someone help me with number 4,5 and 6?

Mathematics

1 answer:

True [87]4 years ago

4 0

4) You know slope-intercept form is y=mx+b. So using these two given points, you can find the slope!

(-8,5) (-3,10) [Use the y1-y2 over x1-x2 formula to solve for slope]

10 - 5 5
--------- = ----- = 1
-3-(-8) 5

Hurray! You got a slope of one. Now substitute this back into your original equation:
y=mx+b --> y=1x+b

Next, we find what our "b" is, or what our y-intercept is:
Using one of the previous points given, substitute them into the new equation:
[I used the point (-3, 10) ]

y=1x+b
10=1(-3)+b SUBSTITUTE
10=-3+b MULTIPLY
10=-3+b
+3 +3 ADD
----------
13=b SIMPLIFY

So, now we have our y-intercept. Use this and plug it into the equation:
y=1x+b --> y=1x+13

y=1x+13 is our final answer.

5) So for perpendicular lines, your slope will be the opposite reciprocal of the original slope. (Ex: Slope is 2, but perpendicular slope is -1/2)

We have the equation y= 3x-1, so find the reciprocal slope!
--> y=-1/3x-1

Good! Now we take our given point, (9, -4) and plug it into the new equation:

y=-1/3x-1
-4=-1/3(9)+b SUBSTITUTE and revert "-1" to "b", for we are trying to find the y- -4=-3+b intercept of our perpendicular equation.
+3 +3 ADD
--------
-1=b SIMPLIFY

So, our final answer is y=-1/3x+(-1)

6) I don't know, sorry! :(

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Answer:

El perímetro de la región impresa es 72 cm y su área es 288 cm².

Step-by-step explanation:

1. Tenemos el perímetro de la hoja de papel:

P₁ = 96 cm = 2l₁ + 2a₁ (1)

Como sabemos el margen superior, inferior, izquierdo y derecho podemos encontrar la relación entre el largo y ancho del rectángulo interno (región impresa) con el largo (l) y ancho (a) del rectángulo externo (hoja de papel):

$l_{2} = l_{1} - (m_{s} + m_{i}) = l_{1} - (3 cm + 2 cm) = l_{1} - 5 cm$ (2)

$a_{2} = a_{1} - (m_{d} + m_{iz}) = a_{1} - (2 cm + 5 cm) = a_{1} - 7 cm$ (3)

El perímetro del rectángulo interno es:

$P_{2} = 2l_{2} + 2a_{2}$ (4)

Introduciendo la ecuación (2) y (3) en (4):

$P_{2} = 2l_{2} + 2a_{2} = 2(l_{1} - 5 cm) + 2(a_{1} - 7 cm) = 2l_{1} + 2a_{1} - 10 cm - 14 cm = 96 cm - 24 cm = 72 cm$

Por lo tanto el perímetro del rectángulo interno (región impresa) es 72 cm.

2. Ahora para encontrar el área rectángulo interno debemos encontrar el largo y ancho del mismo, sabiendo que:

$l_{2} = 2a_{2}$ (5)

Introduciendo (5) en (4):

$P_{2} = 2l_{2} + 2a_{2} = 2*2a_{2} + 2a_{2} = 6a_{2}$

$a_{2} = \frac{P_{2}}{6} = \frac{72 cm}{6} = 12 cm$

$l_{2} = 2a_{2} = 2*12 cm = 24 cm$

Entonces el área es:

$A_{2} = l_{2}*a_{2} = 12 cm*24 cm = 288 cm^{2}$

Por lo tanto el área del rectágulo interno (región impresa) es 288 cm².

Espero que te sea de utilidad!

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3 years ago

. If you know the number of yards for a measurement, you can change that measure to meters by multiplying the number of yards by

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The <u>correct function</u> would be m(y) = 0.9144y.

Explanation:
Let y be the number of yards and m(y) be the number of meters, dependent upon the number of yards.

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4 years ago

Read 2 more answers

A university conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the

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Answer:

The correct answer is c.

Step-by-step explanation:

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P= 6.41÷458×100=1.4%

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3 years ago

Given a graph for the transformation of f(x) in the format g(x) = f(kx), determine the k value. two parabolas open up with f of

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Answer:

I believe that the answer is k=4.

Step-by-step explanation:

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3 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

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Answer:

The interval is [910.053; 959.946]

p-value 0.00596

Decision: Reject null hypothesis.

Step-by-step explanation:

Hello!

You need to make a 95% Confidence Interval for the population mean of scholarship examination scores for the freshman.

It is known to be μ= 900 and the assistant dean wants to test if it changed.

The study variable is:

X: -scholarship examination score of one applicant.

population variance is known as σ²= (180)²

Assuming that the variable has a normal distribution the formula for the interval is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

935 ± 1.96* $\frac{180}{\sqrt{200} }$

The interval is [910.053; 959.946]

To test if the examination scores have changed the hypothesis is:

H₀: μ = 900

H₁: μ ≠ 900

α: 0.05

To use a Confidence Interval the following conditions should be met:

1) Both the test and the interval should be made for the same parameter.

2) The hypothesis has to be two_tailed

3) Confidence level 1 - α and significance level α should be complementary.

To make the decision you have to see if the value given to the population mean in the null hypothesis is contained or not by the interval.

If the value is contained by the interval, you do not reject the null hypothesis.

If the value is not contained by the interval, then the decision is to reject the null hypothesis.

Since 900 is not contained by the 95% Confidence interval [910.05; 959.95], the decision is to reject the null hypothesis. This means that the scholarship examination scores of freshman applications have changed.

To calculate the p-value you have to know the value of the statstic under the null hypothesis:

Z= $\frac{935-900}{\frac{180}{\sqrt{200} } }$ = 2.749 ≅ 2.75

p-value is

P(Z<2.75) + P(Z>2.75)= P(Z<2.75) + (1 - P(Z<2.75))= 0.00298+ (1 - 0.9702= 0.00596

I hope it helps!

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3 years ago