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3 years ago

Please help, pic attached!

Mathematics

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

(y − 9) (y − 2)

Step-by-step explanation:

Factor y^2 − 11y + 18 using the AC method.

(y − 9) (y − 2)

The “AC” method or factoring by grouping is a technique used to factor trinomials. A trinomial is a mathematical expression that consists of three terms.

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Connor rolled a die 6 times. A 2 occurred three times. Why isn't the theoretical probability 3/6 instead of 1/6?

Cerrena [4.2K]

2 occurred 3times

$\\ \sf\longmapsto S=6\times 3=18$

$\\ \sf\longmapsto |S|=18$

Now

$\\ \sf\longmapsto E=No\:of\:times\: 2\:occured$

$\\ \sf\longmapsto |E|=3$

we know

$\boxed{\sf P(E)=\dfrac{|E|}{|S|}}$

$\\ \sf\longmapsto P(E)=\dfrac{3}{18}$

$\\ \sf\longmapsto P(E)=\dfrac{1}{6}$

6 0

3 years ago

Jon recently drove to visit his parents who live 280 miles away. On his way there his average speed was 9 miles per hour faster

Luba_88 [7]

Answer:

11 mph and 20 mph

Step-by-step explanation:

Represent his average speed going by r1 and his average speed returning by r2. We know that r1 = r2 + 9.

Recall that distance = rate times time, so time = distance / rate.

Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).

Time spend returning was (280 mi) / r2.

The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs

Note that there is only one variable here: r2. Find r2, and then from r2, find r1:

Dividing all 3 terms by 14 hrs yields:

20 20

---------- + ----------- = 1

r2 + 9 r2

The LCD here is r2(r2 + 9). Thus, we have:

20r2 (r2 + 9)(r2)

------------------- = 1 or ------------------

(r2 + 9)(r2) (r2 + 9)(r2)

Then 20(r2) = (r2)^2 + 9(r2). This is reducible by dividing all terms by r2:

20 = r2 + 9, or 11 = r2. Then r1 = 11 + 9, or 20.

The two rates were 11 mph and 20 mph.

8 0

3 years ago

Solve for x: <br> dx+fx=8-2dx

Rom4ik [11]

Answer:

fx=8-3dx

Step-by-step explanation:

8 0

3 years ago

Two planes A and B start from the same place and move in different directions making an angle of 50° between them the speed of p

Alika [10]

Answer:

The distance between them is 230.65 miles

Step-by-step explanation:

Here we use the Cosine formula

$a^2 = b^2 + c^2 - 2bc\ cos\ A\\\\a^2 = 200^2 + 300^2 - 2\times 200\times 300cos50^{\circ}\\\\= 40,000 + 90,000 - 120,000cos50^{\circ}\\\\= 130,000 - 120,000 \times 0.64\\\\= 130,000 - 76,800\\\\a^2= 53,200\\\\a = \sqrt{53,200} \\\\= 230.65\ miles\ per\ hour$

Now the distance for one hour is

= 230.65 ÷ 1

= 230.65 miles

4 0

3 years ago

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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 problem, we have that:

$\mu = 591, \sigma = 42$

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{42}$

$Z = 0.21$

$Z = 0.21$ has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{50}} = 5.94$

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{5.94}$

$Z = 1.515$

$Z = 1.515$ has a pvalue of 0.9351

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{100}} = 4.2$

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

$Z = \frac{600 - 591}{4.2}$

$Z = 2.14$

$Z = 2.14$ has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

8 0

3 years ago