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3 years ago

Write an equation for this line

Mathematics

2 answers:

Wewaii [24]3 years ago

5 0

The answer is y=4/-3
Or y=4x-3

Makovka662 [10]3 years ago

3 0

Answer:

y + 3 = 2(x - 4)

Step-by-step explanation:

Use the equation of a line format y - k = m(x - h), where m is the given slope and (h, k) is the given point. Here (h, k) is (4, -3) and m is 2.

Then y + 3 = 2(x - 4) (point-slope form)

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Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering

Alla [95]

Answer:

a) The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

Step-by-step explanation:

a) To approximate this distribution we have to calculate the mean and the standard distribution.

The mean is the proportion p=0.85.

The standard deviation can be calculates as:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{100} }=0.04$

To calculate the probability that Jodi scores 78% or less on a 100-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.04} =-1.75$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) In this case, the number of questions is 250, so the standard deviation needs to be calculated again:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{250} }=0.02$

To calculate the probability that Jodi scores 78% or less on a 250-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.02} =-3.5$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

6 0

3 years ago

The game room of the youth center needs carpet. The room measures 3 feet by 2 feet on the blue print. If the scale on the blue p

sergij07 [2.7K]

The room is 15yd × 10yd
so 15×10=150

the room is 150 square ft

5 0

3 years ago

Read 2 more answers

Please help :(,,,,,,???,,,,,,

nadya68 [22]

Step-by-step explanation:

The first one (f×g)(x) is asking you what is f of x times g of x. This means you multiply the function. You can multiply by the first term (5x) then the second (3) and combine like terms.

${15x}^{2} (5x) = {75x}^{3} \\ 19x(5x) = 95 {x}^{2} \\ 6(5x) = 30x$

${15x}^{2} (3) = 45 {x}^{2} \\ 19x(3) = 57x \\ 6(3) = 18$

Combine like terms

${75x}^{3} + {95x}^{2} + {45x}^{2} + 30x + 57x \\ + 18 = \\ {75x}^{3} + {140x}^{2} + 87x + 18$

The second one asks you to divide.

Factor out f(x):

${15x}^{2} + 19x + 6$

From dividing f(x) /g(x)

$\frac{ {15x}^{2} + 19x + 6 }{5x + 3}$

$\frac{(5x + 3)(3x + 2)}{5x + 3}$

Cancel like terms since (5x+3) is on top and bottom of fraction.

You get 3x+2

5 0

3 years ago

3 photos are printed in 1 minute. How many minuted would it take to print 600 photos

krok68 [10]

Answer:200 minutes

Step-by-step explanation:

1 minute = 60 seconds.

6/3 = 2 (20 seconds) 1 photo every 20 seconds.

600x20 = 12000. seconds.

12000/60 = 200 (200 minutes)

4 0

2 years ago

Read 2 more answers

A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean

Dovator [93]

Answer:

B) 0.0069

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 $s = \frac{\sigma}{\sqrt{n}}$ .