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Hurry Up! I will give brianliest. PLEASE GIVE EXPLANATION or i wont give brainliest

Answers

Answer 1

Answer:

Answer:

x = 19

Step-by-step explanation:

Question: find x such that f(x) = 2/3

Given f(x) = (-1/3) x + 7

equate the value of f(x) to be 2/3

hence,

(2/3) = (-1/3)x + 7 (multiply both sides by 3)

(3) (2/3) = (3) (-1/3)x + (3) 7

2 = -x + 21

x = 21 - 2

x = 19

Answer 2

Answer:

Answer:

x = 19

Step-by-step explanation:

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Anwser choices
A)55
B)83
C)106
D)165

Answers

Answer:

the answer is A

Step-by-step explanation:

A missile uses an internal computer clock to measure time in tenths of a second. The missile guidance system needs the time from launch, in seconds, in order to calculate its distance from the launch site. It obtains this by multiplying the computer clock time by 0.1. So, for example, a reading of 50 tenths of a second is 5 seconds. For one particular missile system, the conversion factor of 0.1 is stored in 8 bits as the binary number 0.000110012.The missile’s internal computer clock shows 200 tenths of a second. Convert this to seconds using your result from Part (i) of this question. Do not round your answer. The missile guidance system now records that it has travelled for this number of seconds.

Answers

Using proportions, it is found that the missile's internal computer clock is of 20 seconds.

According to the information given, this question can be solved by proportions, using a rule of three.

50 tenths of a second is equivalent to 5 seconds, and we want to find the equivalent to 200 tenths of a second, thus, the rule of three is:

50 tenths - 5 seconds

200 tenths - x seconds

Applying cross multiplication:

$50x = 5(200)$

$50x = 1000$

$x = (1000)/(50)$

$x = 20$

The missile's internal computer clock is of 20 seconds.

A similar problem is given at brainly.com/question/24372153

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.(a) What is the probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive? (b) What is the probability that a randomly selected bag contains fewer than 1025 chocolate chips? (c) What proportion of bags contains more than 1200 chocolate chips? (d) What is the percentile rank of a bag that contains 1425 chocolate chips?

Answers

Using the normal distribution, it is found that:

a) There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.

b) There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.

c) 0.3446 = 34.46% of bags contains more than 1200 chocolate chips.

d) The bag is in the 91st percentile.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

It 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, which is the percentile of X.

In this problem:

Mean of 1252 chips, thus $\mu = 1252$ .
Standard deviation of 129 chips, thus $\sigma = 129$ .

Item a:

The probability is the p-value of Z when X = 1500 subtracted by the p-value of Z when X = 1000, thus:

X = 1500:

$Z = (X - \mu)/(\sigma)$

$Z = (1500 - 1252)/(129)$

$Z = 1.92$

$Z = 1.92$ has a p-value of 0.9726.

X = 1000:

$Z = (X - \mu)/(\sigma)$

$Z = (1000 - 1252)/(129)$

$Z = -1.95$

$Z = -1.95$ has a p-value of 0.0256.

Then, 0.9726 - 0.0256 = 0.947

There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.

Item b:

This probability is the p-value of Z when X = 1025, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1025 - 1252)/(129)$

$Z = -1.76$

$Z = -1.76$ has a p-value of 0.0392.

There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.

Item c:

This proportion is 1 subtracted by the p-value of Z when X = 1200, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1200 - 1252)/(129)$

$Z = -0.4$

$Z = -0.4$ has a p-value of 0.3446.

0.3446 = 34.46% of bags contains more than 1200 chocolate chips.

Item d:

This percentile is the p-value of Z when X = 1425, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1425 - 1252)/(129)$

$Z = 1.34$

$Z = 1.34$ has a p-value of 0.91.

The bag is in the 91st percentile.

A similar problem is given at brainly.com/question/13680644

Final answer:

The probability and percentiles can be found using the z-score formula and then looking these z-scores up in the standard normal distribution table.

Explanation:

In this question, we're dealing with a normal distribution. For a normal distribution, probabilities can be calculated using the z-score formula which is given by Z = (X - μ) / σ, where X is the value from which you want to find the probability, μ is the mean and σ is the standard deviation.

(a) To find the probability that a randomly selected bag has between 1000 and 1500 chocolate chips, we need to find the z-scores for both 1000 and 1500 and then find the area between these two z-scores in the standard normal distribution table. (b) To find the probability that a randomly selected bag has less than 1025 chips, we find the z-score for 1025 and find the area to the left of it in the standard normal distribution table. (c) To find the proportion of bags containing more than 1200 chips, we find the z-score for 1200 and then find the area to the right of it in the standard normal distribution table. (d) Percentile rank can be found by finding the z-score for 1425 chips, and then finding the corresponding percentile in the standard normal distribution table.

Learn more about Normal Distribution here:

brainly.com/question/34741155

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There are 250 students in the senior class at Stats High School. Some are involved in clubs and some have part-time jobs. 130 students have a part-time job (some of these students may be in a club). 110 students are involved in a club (some of these students may have a job). 100 students are NOT involved in a club and do NOT have a part-time job1. Draw a Venn diagram to illustrate the data.

Answers

Answer: ( i had a problem with this question too and i looked it up for a tutorial and i saw that some guy replied random things for points, so after i found the explanation, i came back here to give you a proper answer.)

Write 100 outside the circles to represent the students who do not have a job and are not in a club.

There are a total of 250 students, and 100 of them are not in the circles. So the remaining 150 must be included in the job and club circles.

The total number in the job subset is 130, and the total number in the club subset is 110. Together, this is 240 students, but there is only room for 150.

This means that 90 of the students are included in the intersection (both job and club).

The total number of students with jobs is 130, but 90 are already included in the intersection. Therefore, the difference of 40 students only have jobs.

The total number of students in a club is 110, but 90 have both jobs and attend a club. So the remaining 20 students must only attend a club.

Is 4:3 and 3:4 equivalent