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3 years ago

Determine whether y = 4x+5 and y= 1/4x-2 are perpendicular

Mathematics

1 answer:

Bumek [7]3 years ago

3 0

Answer:

No, they are not perpendicular

Step-by-step explanation:

Perpendicular lines will have opposite reciprocal slopes, meaning that the slopes will have opposite signs and be the reciprocals of each other.

We can see that the slopes of the lines are not opposite reciprocals.

4 and 1/4 are not opposite reciprocals because they do not have opposite signs.

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Each baby dinosaur made 15 paintings and each adult dinasour made 7 paintings. The entire herd made 208 paintings in total, and

Serggg [28]

Answer:

there are 6 baby dinos and 2

Step-by-step explanation:

6 0

3 years ago

Reasoning Kareem cannot decide which of two washing machines to buy. The selling price of each is $720. The first is marked down

Marina CMI [18]

Answer:

store one

Step-by-step explanation:

Store One:

$720(0.50)=$360

$720-$360=$360

Store Two:

$720(0.30)=$216

$720-$216=$504

$504(0.20)=$100.8

$504-$100.8=$403.2

5 0

3 years ago

FRAME AN EQUATION AND SOLVE IT. NEED HELP ASAP I only have 5 minutes!

lawyer [7]

Answer:

The answer to your question is 125

Step-by-step explanation:

x = 25 X 5= 125.

The required number is 125.

I hope this helps and have a good day!

3 0

3 years ago

Find the point-slope equation for

mars1129 [50]

Answer:

y-24=m(x--4)

Step-by-step explanation:

point slope form:

y-y₁=m(x-x₁)

1) find the slope or "m" first:

slope formula: (y2-y1)/(x2-x1)

(-53-24)/7+4

-77/11=-7

slope=-7

y-y₁=-7(x-x₁)

y1=24

x1=-4

y-24=m(x--4)

hope this helps!

8 0

3 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago