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3 years ago

16t-2(2t+5)=14 7th grade T=? plz helpppp

2 answers:

8 0

16t -2(2t+5) =14

distribute the parenthesis first:

16t-4t-10 = 14

combine like terms:

12t - 10 = 14

add 10 to both sides:

12t = 24

divide both sides by 12

t = 24/12

t = 2

timofeeve [1]3 years ago

3 0

Not very good at math but i believe the answer is16t -2(2t+5) =14

distribute the parenthesis first:

16t-4t-10 = 14

combine like terms:

12t - 10 = 14

add 10 to both sides:

12t = 24

divide both sides by 12

t = 24/12

t = 2

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Help me with the right answer

Nadya [2.5K]

Answer:

x = 3

y = 1

Step-by-step explanation:

9x - y = 26 ---------------(I)

9x + y = 28 ---------------(II)

9x - y = 26

9x = 26 + y

Substitute 9x = 26 + y in equation (II)

26 + y + y = 28 {add like terms}

26 + 2y = 28 {subtract 26 from both sides}

2y = 28 - 26

2y = 2 {divide both sides by 2}

y = 2/2

y = 1

Substitute y =1 in equation (I)

9x - 1 = 26

9x = 26 + 1

9x = 27

x = 27/9

x = 3

y = 1

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3 years ago

At Pike Place Fish Market in Seattle, customers can purchase a variety of different types of seafood. One type of seafood sold a

allsm [11]

Answer:

$z=1.28$

And if we solve for a we got

$a=3.6 +1.28*0.8=4.624$

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

C. 4.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.6,0.8)$

Where $\mu=3.6$ and $\sigma=0.8$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=3.6 +1.28*0.8=4.624$

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

C. 4.64

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3 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$