1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anestetic [448]
3 years ago
5

A square picture has a 1-in. frame around it. If the area of the frame alone is 36 in.^2, what is the area of the picture in squ

are inches?
Mathematics
1 answer:
Oliga [24]3 years ago
8 0

Answer:

The area of the picture is

Step-by-step explanation:

Let

x-----> the length side of the square picture

we know that

The area of the picture plus the frame minus the area of the picture is equal to the area of the frame alone

The area of the picture is

You might be interested in
Can anyone help with the last problem about The Difference Quotient of a Function
Marysya12 [62]

Answer:

tbh i really dont know

Step-by-step explanation:

3 0
2 years ago
I need help with this asap!!! pls​
Len [333]

Answer:

ok so the answers are for all of them number 1-6 sorry

Step-by-step explanation:

1) unite rate

2) constant rate

3) proportional relationship

4) ordered pair

5) greater than

6) less than

hope this helps girlyy<33

4 0
3 years ago
Suppose that we are testing a coin to see if it is fair, so our hypotheses are: H0: p = 0.5 vs Ha: p ≠ 0.5. In each of (a) and (
aleksley [76]

Answer:

H_0: p = 0.5\\H_a: p \neq 0.5

a. We get 56 heads out of 100 tosses.

We will use one sample proportion test  

x = 56

n = 100

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{56}{100}

\widehat{p}=0.56

Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

                                       =\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}

                                       =1.2

refer the z table for p value

p value = 0.8849

a.  We get 560 heads out of 1000 tosses.

We will use one sample proportion test  

x = 560

n = 1000

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{560}{1000}

\widehat{p}=0.56

Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

                                       =\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}

                                       =3.794

refer the z table for p value

p value = .000148

p value of part B is less than Part A because part B have 10 times the number the tosses.

6 0
2 years ago
If i= the square root of -1 what is the value of i^3
BARSIC [14]

i =√-1

i^2 = -1

so

i^3

= i^2 * i

=  -1√-1

= - √-1

5 0
3 years ago
Mystery Boxes: Breakout Rooms
ollegr [7]

Answer:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

Step-by-step explanation:

Given

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}

Required

Fill in the box

From the question, the range is:

Range = 60

Range is calculated as:

Range =  Highest - Least

From the box, we have:

Least = 1

So:

60 = Highest  - 1

Highest = 60 +1

Highest = 61

The box, becomes:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question:

IQR = 20 --- interquartile range

This is calculated as:

IQR = Q_3 - Q_1

Q_3 is the median of the upper half while Q_1 is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}

<u>Upper half</u>

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

Median = \frac{N + 1}{2}th

Where N = 7

So, we have:

Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th

So,

Q_3 = 4th item of the upper halves

Q_1= 4th item of the lower halves

From the upper halves

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

<u></u>

We have:

Q_3 = 32

Q_1 can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

IQR = Q_3 - Q_1

Where Q_3 = 32 and IQR = 20

So:

20 = 32 - Q_1

Q_1 = 32 - 20

Q_1 = 12

So, the lower half becomes:

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}

From this, the updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question, the median is:

Median = 22 and N = 14

To calculate the median, we make use of:

Median = \frac{N + 1}{2}th

Median = \frac{14 + 1}{2}th

Median = \frac{15}{2}th

Median = 7.5th

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

Mode = 18

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

7th = 18

The 8th item is calculated as thus:

Median = \frac{1}{2}(7th + 8th)

22= \frac{1}{2}(18 + 8th)

Multiply through by 2

44 = 18 + 8th

8th = 44 - 18

8th = 26

The updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question.

Mean = 26

Mean is calculated as:

Mean = \frac{\sum x}{n}

So, we have:

26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}

Collect like terms

26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}

26= \frac{ 2nd + 12th+304}{14}

Multiply through by 14

14 * 26= 2nd + 12th+304

364= 2nd + 12th+304

This gives:

2nd + 12th = 364 - 304

2nd + 12th = 60

From the updated box,

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

We know that:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

2nd = 3

12th = 57

i.e.

2nd + 12th = 60

3 + 57 = 60

So, the complete box is:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

6 0
2 years ago
Other questions:
  • To earn money, George types papers for college students. For regular term papers, he charges by the page: $1.50 each. For scient
    7·1 answer
  • mrs. Walker increase the amount of fruit she eats everyday from 25 to 30 by what percentage did mrs. Walker increase the amount
    10·2 answers
  • How do you find the gcf of 2 numbers
    9·2 answers
  • X-6/2=2x/7 solve the equation​
    5·1 answer
  • Find g(x) if it is known that g(2t)=8t−1.
    14·1 answer
  • You are planning on making 250 liters of chili for a big fundraiser you are throwing. You want the
    6·1 answer
  • How do I Convert Mixed Numbers to Improper Fractions
    5·2 answers
  • FREE BRAINLIEST!!!!!!
    15·2 answers
  • 35POINTS PLZ HELP FAST
    15·1 answer
  • A rectangle with an area of 5/8 ft^2 dilated by a factor of 8.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!