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Thepotemich [5.8K]
2 years ago
6

The length of a rectangle is three times its width.

Mathematics
1 answer:
Igoryamba2 years ago
8 0
96 divided by 4 is 24
3times 24 is72
And 72 is 3 times 24
And 72plus 24 is 96
So your answer is length 72
Wide24
You might be interested in
how much would $150 invested at 8% interest compounded anually be worth after 13 years? A(t)=P(1+r/n)^
PIT_PIT [208]
\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$150\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &13
\end{cases}
\\\\\\
A=150\left(1+\frac{0.08}{1}\right)^{1\cdot 13}\implies A=150(1.08)^{13}\implies A\approx 407.9435589
3 0
3 years ago
Which is the graph of x – y = 1?
Jet001 [13]

Answer:

its the second graph

Step-by-step explanation:

you could use this website to find out graph

desmos calculator

5 0
3 years ago
7. The manager of the soda shop decides that because you are a good customer, you may use a flavor as many times as you’d like.
Black_prince [1.1K]

The different possible mixtures can you make is 1716 .

<u>Step-by-step explanation:</u>

Given that,

  • The total number of flavors available in the shop = 13 flavors
  • The number of flavors you need to choose = 6 flavors

Here, we have to use  the formula for combination,

nCr = n! / r! × (n-r) !

where,

  • n is the total number of flavors
  • r is the number of flavors you need to choose

⇒ 13C6 = 13 ! / 6! × (13-6)!

⇒ 13! / (6! × 7! )

⇒ 13 × 12 × 11 × 10 × 9 × 8 × 7! / (6! × 7! )

⇒ 13 × 12 × 11 × 10 × 9 × 8 / 6!

⇒ 13 × 12 × 11 × 10 × 9 × 8 / 6 × 5 × 4 × 3 × 2 × 1

⇒ 1716

Therefore, you can 1716 different possible mixtures.

3 0
3 years ago
The data set represents the total number of pencils each student in a class needs to sharpen.
Firdavs [7]

Given:

The data set is:

0, 1, 1, 1, 2, 3, 4, 4, 6, 6, 9

To find:

The correct statement that represents the box plot of the data correctly.

Solution:

We have,

0, 1, 1, 1, 2, 3, 4, 4, 6, 6, 9

Divide the data in 2 equal parts.

(0, 1, 1, 1, 2), 3, (4, 4, 6, 6, 9)

Divide each parenthesis in 2 equal parts.

(0, 1), 1, (1, 2), 3, (4, 4), 6, (6, 9)

Here,

Minimum value = 0

First quartile = 1

Median = 3

Third quartile = 6

Maximum value = 9

A number line goes from 0 to 10. The whiskers range from 0 to 9, and the box ranges from 1 to 6. A line divides the box at 3.

Therefore, the correct option is D.

5 0
3 years ago
Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut str
sergeinik [125]

Answer:

Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the  left-tailed.

{0,1,2,3,4,5} ;  

{15,16,17,18,19,20} is a right-tailed test and not appropriate for H0:

{ 0,1,2,3,17,18,19,20} is two-tailed and not appropriate for H0:

b)

Does the region specify a level .05 test? No

 

P = proportion who prefer gut strings to nylon

P = X /20

Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) <  < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20   < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20   < 0.316

Reject if x   < (0.316)(20) = 6.32

{0,1,2,3,4,5,6} is the region for the best level 0.05 test

c)

According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

add: 0.9984 --  proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

add: 1.0000  probability of a type II error

d)

P( x <= 13) =  

0  0.000001  

1  0.000019  

2  0.000181  

3  0.001087  

4  0.004621  

5  0.014786  

6  0.036964  

7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

add: 0.9423 < 0.10 ,  H0 cannot be rejected

5 0
3 years ago
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