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3 years ago

I need help on this word problem!

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Width be X, length is X+15

area of garden = x(x+15)
area of sidewalk = x(x+15) - (x-4)(x+11) = 632

X^2 + 15X - X^2 - 7X + 44 = 632
8X = 632-44 =588

X= 73.5 >> WIDTH and LENGTH = 88.5

HOPE IT HELPED

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The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n + 2} = F_{n + 1} + F_n$. Find the remainder when $F_{1999}$ is di

zavuch27 [327]

Answer:

The remainder is 1

Step-by-step explanation:

Given the Fibonacci sequence

F_1 = F_2 = 1, and

F_(n + 2) = F_(n + 1) + F_n

We want to find the remainder when F_(1999) is divided by 5.

Let us write the first 20 numbers of the sequence in (mod 5). They are

F_1 = 1,

F_2 = 1,

F_3 = 2,

F_4 = 3,

F_5 = 5 = 0 (mod 5),

F_6 = 3,

F_7 = 3,

F_8 = 1

F_(9) = 4

F_(10) = 0

F_(11) = 4

F_(12) = 4

F_(13) = 3

F_(14) = 2

F_(15) = 0

F_(16) = 2

F_(17) = 2

F_(18) = 4

F_(19) = 1

F_(20) = 0

We have: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0

Now, 1999 = 19(mod 20)

The 19th number in the sequence is 1.

So, the remainder is 1.

6 0

3 years ago

Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size

julsineya [31]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641$

And rounded up we have that n=16641

Step-by-step explanation:

Previous concepts

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

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

Solution to the 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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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 $ME =\pm 0.01$ 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)

For this case we can use as estimator for the proportion $\hat p =0.5$ , since we don't have any other previous info. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641$

And rounded up we have that n=16641

8 0

3 years ago

How do you solve this?

Law Incorporation [45]

Find the soluiton

2x+2y=16
3x-y=4

x+y=8
<u>3x-y=4+</u>
4x=12
x=3
3(3)-y=4
9-y=4
y=5
(3,5)
test them to see if get true statment
obviously the first equatons of 1,2,4 work
1 doesn't work
2, works
4. doesn't work

2 is the same as the first except both euations are just doubled

answer is 2

6 0

3 years ago

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

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