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3 years ago

How to express a polynomial sum in simplest form?

Mathematics

1 answer:

klemol [59]3 years ago

6 0

Answer:

Polynomials can be simplified by using the distributive property to distribute the term on the outside of the parentheses by multiplying it by everything inside the parentheses. You can simplify polynomials by using FOIL to multiply binomials times binomials.

Step-by-step explanation:

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Solve the simultaneous equation: 5 x + 2 y = 25 5 x − y = 10

Liono4ka [1.6K]

Answer:

(x, y) = (3, 5)

Step-by-step explanation:

$5x+2y=25\\5x-y=10$

Solving by elimination here again, there are 2 good options available. Either multiply the whole bottom equation by -1 to cancel the x, or by 2 to cancel the y. I'll do the latter:

$5x+2y=25\\\\2(5x-y=10)\\10x-2y=20$

Add from top to bottom:

$\rightarrow (5x+10x)+(2y-2y)=25+20\\\rightarrow 15x=45\\\rightarrow x=3$

Now, with the value of x, solve for y in either of the equations. I'll choose the second one here:

$\rightarrow 5x-y=10\\\rightarrow 5(3)-y=10\\\rightarrow 15-y=10\\\rightarrow -y=-5\\\rightarrow y=5$

(x, y) = (3, 5)

4 0

2 years ago

Read 2 more answers

The area of the base b is the area of a square with sides of length 230 m. so, v =

goldfiish [28.3K]

Not completely sure what you are asking, but the area of a square with the side lengths of 230m would be 230*230=52,900.

4 0

3 years ago

Suppose there are signs on the doors to two rooms. The sign on the first door reads "In this room there is a lady, and in the ot

9966 [12]

Answer:

The lady is on the second door

Step-by-step explanation:

we have that

The sign on the first door reads "In this room there is a lady, and in the other one there is a tiger"

The sign on the second door reads "In one of these rooms, there is a lady, and in one of them there is a tiger."

The sign on the second door is true

The sign on the first door is true or false

Since one of these signs is true and the other is false, the sign in the first door must be false

therefore

The lady is on the second door

5 0

3 years ago

What two numbers add up to 6.5 and multiply to .64

melomori [17]

You multiply by 10 or more

5 0

3 years ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago