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hjlf
3 years ago
12

Find the discounted price of the teddy bear which has a regular price of $450.

Mathematics
1 answer:
GuDViN [60]3 years ago
6 0
You will pay $360 for a item with original price of $450 when discounted 20%. 450x0.2=90.00 then 450-90=360
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The degree of the polynomial function f(x) is 3. The roots of the equation f(x)=0 are −4 , 0, and 2. Which graph could be the gr
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We know that the polynomial function is of degree 3, and that its roots are -4, 0, 2.

With this data we can write a generic equation for the function:

f (x) = bx (x + 4) (x-2)

Since the function is of degree 3 and cuts the axis at x = 0, then it has rotational symmetry with respect to the origin.

The graph of the function can be of two main forms, based on the value of the coefficient b.

If b is positive then the function grows from y = -infinite and cuts the x-axis for the first time in -4. Then it decreases, cuts at x = 0 and begins to grow again cutting the x-axis for the third time at x = 2. and continues to grow until y = infnit


If b is negative, then the function decreases from y = infinity and cuts the x-axis for the first time in -4. Then it grows, cuts at x = 0 and begins to decrease again by cutting the x-axis for the third time at x = 2, and continues to decrease until y = -infnit.


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3 years ago
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Find an acute angle whose sine is 0.52
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3 years ago
The following random sample was selected from a normal distribution: 4.5, 6.4, 2.3, 1.8, 5.3, then the 95% confidence interval t
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Answer:

the 95% confidence interval to estimate the population mean is between  1.62 and 6.50

Step-by-step explanation:

given data

distribution = 4.5, 6.4, 2.3, 1.8, 5.3

so n = 5

confidence interval = 95%

to find out

the population mean is between

solution

first we calculate the mean i.e.

mean =  \frac{1}{n}\sum_{i=1}^{n}x(i)

mean =  4.5+ 6.4+ 2.3+ 1.8+ 5.3 / 5

mean = 4.06

now we calculate the standard deviation i.e.

standard deviation =   \sqrt{\frac{1}{n-1}\sum (x(i)-mean)^2}

standard deviation =   \sqrt{\frac{1}{5-1}\sum (x(i)-mean)^2}

standard deviation =   \sqrt{\frac{1}{5-1} (4.5-5)^2+(6.4-5)^2 +(2.3-5)^2+(1.8-5)^2+(5.3-5)^2}

standard deviation =   \sqrt{\frac{1}{4} (4.5-5)^2+(6.4-5)^2 +(2.3-5)^2+(1.8-5)^2+(5.3-5)^2}

standard deviation =  2.226544

so 95 % confidence interval is i.e. mean +/- t(5) * standard deviation/ \sqrt{n}

here t(5) will be 2.45

so

95 % confidence interval=  4.06 +/- 2.45 * 2.226 / \sqrt{5}

95 % confidence interval= 1.62 and 6.50

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Answer:

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