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3 years ago

What is 10*10/11 simplified to a mixed number

Mathematics

2 answers:

g100num [7]3 years ago

8 0

Answer:

9 1/11

Step-by-step explanation:

10*10= 100

100/11

9 and 1/11

Sunny_sXe [5.5K]3 years ago

8 0

Answer:

it is already a mixed number

Step-by-step explanation:

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Make a conjecture abouth the following statement: Angle ABC and angle CBD are adjacent.

zvonat [6]

The measure of Angle ABD = measure of Angle ABC + measure of Angle CBD

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2 years ago

In a certain clothing store, 6 shirts and 3 ties cost $79.50, and 3 shirts and 2 ties cost $41. Determine the cost of each tie.

Genrish500 [490]

6x + 3y = 79.5
3x + 2y = 41 y = -3/2x + 41/2 y = -1.5x + 20.5

6x + 3(-1.5x + 20.5) = 79.5
6x - 4.5x + 61.5 = 79.5
1.5x + 61.5 = 79.5
1.5x = 18
x = 12

Each shirt costs $12.
Now insert that number into the x to find the cost of each tie.

3(12) + 2y = 41
36 + 2y = 41
2y = 5
y = 2.5

Each tie costs $2.50.

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3 years ago

Rita bought a pair of jeans for $89. The sales-tax rate is 11 percent. What is the total amount she paid for the jeans?

andre [41]

Total amount paid by Rita is $98.79

Step-by-step explanation:

Step 1: Given sales tax = 11% and cost of jeans = $89
Step 2: Calculate amount of sales tax = 11% of 89 = 11/100 × 89 = $9.79
Step 3: Calculate the total amount

Total Amount = 89 + 9.79 = $98.79

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3 years ago

Jackie is creating a flower bed in the shape of a triangle what could be the possible measurement in feet

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3 years ago

The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ou

Setler [38]

Answer:

Step-by-step explanation:

(a)

The distribution of X is Normal Distribution with mean $= \mu =17$ and Variance $= \sigma^{2} = 16 \ i.e., X \sim N (17, 16),$

(b)

The distribution of $\bar{x}$ is Normal Distribution with mean $= \mu =17$ and Variance = $\sigma^{2}/n = 16/15= 1.0667$ .i.e., $\bar{x}\sim N(17,1.0667)$

To find P(15.5 < X < 18):

Case 1: For X from 15.5 to mid value:

Z = (15.5 - 17)/4 = - 0.375

Table of Area Under Standard Normal Curve gives area = 0.1480

Case 2: For X from mid value to 18:

Z = (18 - 17)/4 = 0.25

Table of Area Under Standard Normal Curve gives area = 0.0987

So,

P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467

So,

Answer is:

0.2467

(d)

$SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}$

= 1.0328

To find $P(15.5 < \bar{x}< 18):$

Case 1: For $\bar{x}$ from 15.5 to mid value:

Z = (15.5 - 17)/1.0328 = - 1.4524

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 18:

Z = (18 - 17)/1.0328 = 0.9682

Table of Area Under Standard Normal Curve gives area = 0.3340

So,

$P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605$

So,

Answer is:

0.7605

(e)

Correct option:

because Population SD is provided.

(f)

(i)

Q1 is given by:

$- 0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034

So,

Q1 = 16.3034

(ii)

Q3 is given by:

$0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966

So,

Q3= 17.6966

(iii)

IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932

Answers are:

Q1 = 16.3034 ounces

Q3 = 17.6966 Ounces

IQR = 1.3932 Ounces

8 0

3 years ago