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2 years ago

Write a quadratic function f whose only zero is -9

Mathematics

2 answers:

frutty [35]2 years ago

7 0

Answer:

x² + 18x + 81 = 0

Step-by-step explanation:

For the zero to -9, the equation has to be x + 9.

x = -9

x + 9 = 0

Since the only zero is -9, square the equation. Expand to get the function

(x + 9)² = 0

x² + 18x + 81 = 0

Radda [10]2 years ago

7 0

Answer:

$x^{2}+18x+81=0$

Step-by-step explanation:

For it to have zero of -9 it is (x+9)^2. Expand that out and you get the answer.

You might be interested in

One nanometer equals about

Orlov [11]

Answer:

$1\ \text{nanometer}=4\times 10^8$

Step-by-step explanation:

Given : One nanometer equals about 0.00000004 inches.

To find : When writing this decimal as a single-digit integer multiplied by a power of ten, the single digit integer is and the power of ten ?

Solution :

$1\ \text{nanometer}=0.00000004$

$1\ \text{nanometer}=\frac{00000004}{100000000}$

$1\ \text{nanometer}=4\times 10^8$

Therefore, the decimal as a single-digit integer multiplied by a power of ten is $1\ \text{nanometer}=4\times 10^8$

4 0

2 years ago

preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. It was found tha

VladimirAG [237]

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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 population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

And rounded up we have that n=369

8 0

2 years ago

Suppose a1=12−12,a2=23−13,a3=34−14,a4=45−15,a5=56−16. a) Find an explicit formula for an: . b) Determine whether the sequence is

lorasvet [3.4K]

I'm going to assume you meant to write fractions (because if $a_n$ are all non-negative integers, the series would clearly diverge), so that

$a_1=\dfrac12-\dfrac12$

$a_2=\dfrac23-\dfrac13$

$a_3=\dfrac34-\dfrac14$

and so on.

a. If the pattern continues as above, we would have the general term

$a_n=\dfrac n{n+1}-\dfrac1{n+1}=\dfrac{n-1}{n+1}$

b. Note that we can write $a_n$ as

$a_n=\dfrac{n-1}{n+1}=\dfrac{n+1-2}{n+1}=1-\dfrac2{n+1}$

The series diverges by comparison to the divergent series

$\displaystyle\sum_{n=1}^\infty\frac1n$

4 0

2 years ago

The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that P(82 − q &l

mezya [45]

P(82 - q < x < 82 + q) = 0.44
P(x < 82 + q) - P(82 - q) = 0.44
P(z < (82 + q - 82)/7.4 - P(z < (82 - q - 82)/7.4) = 0.44
P(z < q/7.4) - P(z < -q/7.4) = 0.44
P(z < q/7.4) - (1 - P(z < q/7.4) = 0.44
P(z < q/7.4) - 1 + P(z < q/7.4) = 0.44
2P(z < q/7.4) - 1 = 0.44
2P(z < q/7.4) = 1.44
P(z < q/7.4) = 0.72
P(z < q/7.4) = P(z < 0.583)
q/7.4 = 0.583
q = 0.583 x 7.4 = 4.31

8 0

2 years ago

Let d(t) be the total number of miles Joanna has cycled, and let t represent the number of hours after stopping for a break duri

katrin [286]

For this case we have the following equation:
$d (t) = 12t + 20
$
The variables are:
t: time in hours
d (t): distance traveled in miles.
We evaluate the distance traveled after 4 hours.
So we have that for t = 4:
$d (4) = 12 (4) +20

d (4) = 48 + 20

d (4) = 68$
Answer:
d (4) = 68
This means that after 4 hours Joanna cycled for 68 miles total

6 0

2 years ago

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