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Brilliant_brown [7]
3 years ago
7

What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x/4+5 ?

Mathematics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

y = -4x+15

Step-by-step explanation:

y=x/4+5 may be presented as y = 1/4 x + 5, where a=1/4, b=5

swapping numerator and denominator and inserting an opposite sign in a=1/4, we have:

1/4 --------> 4/1 -------> -4/1 = -4

y = -4x+n

Point (2,7) means that x=2, y=7 substitute it to the equation y=-4x+n, calculate "n":

7=-4*2+n ------>    7=-8+n ------->   7+8 = n -------> 15=n

substituting n=15 to the equation y=-4x+n, we get finally y=-4x+15


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8 0
3 years ago
The table below represents the equation: y = 1.5x – 2 If you completed this table, would each input have exactly one output? yes
stealth61 [152]

Answer:

Step-by-step explanation: y=1.5(3)-2

y=2.5

7 0
3 years ago
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A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥
gregori [183]

Answer:

B) (35, 260)

Step-by-step explanation:

A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥ 1∕5w2, where d is dosage in milligrams and w is the dog's weight in pounds. Which of the following ordered pairs gives an effective dosage of antibiotics for a 35-pound dog?

A) (35, 240)

B) (35, 260)

C) (260, 35)

D) (240, 35)

Ordered pairs is composed of pairs, usually an x coordinate and a y coordinate. It refers to a location of a point on the coordinate. It matches numbers to functions or relations.

Given the relation between d is dosage in milligrams and w is the dog's weight in pounds as d ≥ 1∕5w²

For a 35 pound dog (i.e w = 35 pound). The dosage is given as:

d ≥ 1∕5(35)² ≥ 245 milligrams.

For an ordered pair (x, y), x is the independent variable (input) and y is the dependent variable (output).

The dog weight is the independent variable and the dosage is the dependent variable.

From the ordered pairs, the best option is (35, 260) because 260 ≥ 240

6 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
4 years ago
Give that m/_ y=39, find the measures of angles a and b
Amanda [17]

Answer:

  ∠a = 39°

  ∠b = 129°

Step-by-step explanation:

Vertical angles are congruent. An exterior angle of a triangle is equal to the sum of the remote interior angles.

<h3>Application</h3>

Angle y and angle 'a' are vertical angles, so congruent.

  ∠a = ∠y = 39°

The angle marked 'b' is an exterior angle to the triangle. Its remote interior angles are 'a' and the one marked 90°.

  ∠b = ∠a +90° = 39° +90°

  ∠b = 129°

8 0
1 year ago
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