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Nina [5.8K]
3 years ago
9

CAN SOMEONE PLEASE HELP ME SOLVE THIS I WILL GIVE U BRAINIEST PLEASE AND THANK YOU.!

Mathematics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

Step-by-step explanation:

SIMPLE 30 X 30

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The National Science Foundation (NSF) offers grants to students who are interested in pursuing a college education. The student
AlexFokin [52]

Answer:31,606 explanation: took edg test

8 0
1 year ago
for the school play adult tickets cost 4$ and children tickets cost 2$ natalie is working at the ticket counter and just sold 20
lilavasa [31]
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be

4x + 2y = 20

There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.

When x=0,
4(0) + 2y = 20
y = 10

When x=1,
4(1) + 2y = 20
y = 8

When x=2,
4(2) + 2y = 20
y = 6

When x=3,
4(3) + 2y= 20
y = 4

When x = 4,
4(4) + 2y = 20
y = 2

When x = 5,
4(5) + 2y = 20
y = 0

When x = 6,
4(6) + 2y = 20
y = -2

A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:

   Number of adult tickets             Number of children tickets
                  0                                                   10
                  1                                                    8
                  2                                                    6
                  3                                                    4
                  4                                                    2
                  5                                                    0




6 0
3 years ago
Write the equation of the line in fully simplified slope intercept form
yan [13]

The equation of the given line is y = 5/6x - 2

The standard equation of a line in slope-intercept form is expressed as

y = mx + b

m is the slope of the line

b is the y-intercept

Using the coordinate point (0, -2) and (-6, 3)

Slope m = 3-(-2)/-6-0

m = 5/6

Since the line cuts the y-axis at y = -2, hence the y-intercept is -2.

Find the equation of a line;

y = mx + b

y = 5/6x + (-2)

y = 5/6x - 2

Hence the equation of the given line is y = 5/6x - 2

Learn more on equation of a line here: brainly.com/question/7098341

8 0
2 years ago
Find the number that we can place in the box, so that the resulting quadratic is the square of a binomial.
Arturiano [62]

We need a number to complete

(x+c)^2=x^2+2xc+c^2

Since we have

2xc=\dfrac{4}{5}x \iff 2c=\dfrac{4}{5}\iff c=\dfrac{4}{10}=\dfrac{2}{5}

And so the expanded square is

\left( x+\dfrac{2}{5}\right)^2=x^2+\dfrac{4}{5}x+\dfrac{4}{25}

6 0
3 years ago
Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.9 per year. a. Find t
VLD [36.1K]

Answer:

a) 9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b) 4.284 years are expected to have 4 ​hurricanes.

c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

6.9 per year.

This means that \mu = 6.9

a. Find the probability​ that, in a​ year, there will be 4 hurricanes.

This is P(X = 4).

So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}

P(X = 4) = 0.0952

9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b. In a 45​-year ​period, how many years are expected to have 4 ​hurricanes?

For each year, the probability is 0.0952.

Multiplying by 45

45*0.0952 = 4.284.

4.284 years are expected to have 4 ​hurricanes.

c. How does the result from part​ (b) compare to a recent period of 45 years in which 4 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?

The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

5 0
3 years ago
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