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3 years ago

Adding fractions and mixed numbers. 5 1/3+12 1/2

Mathematics

2 answers:

Viefleur [7K]3 years ago

6 0

Answer:17 5/6

Step-by-step explanation:

lawyer [7]3 years ago

5 0

The answer is 17 5/6 I hope this helps you

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How do I create an equivalent expression that includes parentheses so that the value of the expression is 2 for 3+8-4x2-12

natka813 [3]

(3+8-4)×2-12
3+8=11
11-4=7
7×2=14
14-12=2

5 0

3 years ago

An architect makes a model of a new house. The model shows a tile patio in the backyard. In the model, each tile has length

aleksley [76]

Answer:

Step-by-step explanation:

3 0

3 years ago

I NEED HELP ON THIS PLZZZ HELPP

adell [148]

Answer:

$76 \: {in}^{2}$

Step-by-step explanation:

Area of the figure = Are of square with side 8 in + 2 times the area of one triangle with base (8 - 5 = 3) 3 in and height 4 in

$= {(8)}^{2} + 2 \times \frac{1}{2} \times 3 \times 4 \\ \\ = 64 + 12 \\ \\ = 76 \: {in}^{2}$

3 0

3 years ago

Don't get this question

solniwko [45]

Answer:

17 and 18

Step-by-step explanation:

Writing every 2 digit number gives us :

10-99 including 10 and 99

For the first one we are looking for numbers where the number 3 appears only once so 33 would be invalid.

13,23,30,31,32,34,35,36,37,38,39,43,53,63,73,83,93

There are 17 2-digit numbers that have 3 exactly once

For the second one we are looking for number where the number 3 appears a minimum of once so 33 would be valid :

13,23,30,31,32,33,34,35,36,37,38,39,43,53,63,73,83,93

There are 18 2-digit numbers that have 3 at least once

Hope this helped and have a good day

5 0

2 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

5 0

3 years ago