What is this this sos 3x-y=3? (x,y)=?

Lesson 4-3 Dividing Mixed Numbers Practice and Problem Solving: A/B

muminat

Answer:

<h3>Reciprocals are: </h3>

1) 10 1/2 = 21/2

Reciprocal will be 2/21

2) 6 3/7 = 45/7

Reciprocal will be 7/45

3) 2 8/9 = 26/9

Reciprocal will be 9/26

4) 15 1/4 = 61/4

Reciprocal will be 4/61

5) 9 2/3 = 29/3

Reciprocal will be 3/29

6) 7 5/8 = 61/8

Reciprocal will be 8/61

<h3>Divide</h3>

7) 8/10 divide by 1 5/6

8/10 divide by 11/6

or 8/10 * 6/11

4/5* 6/11

24/55

8) 2 divides by 1 6/7

2 divides by 13/7

2 * 7/13

14/13

1 1/13

9) 3 3/5 divides by 2 1/4

18/5 divides by 9/4

18/5 * 4/9

8/5

1 3/5

10) 4 1/2 divides by 2 3/8

9/2 divides by 19/8

9/2 * 8/19

36/19

1 17/19

11) 5 5/6 divides by 3 1/6

35/6 divides by 19/6

35/6 * 6/19

35/19

1 16/19

12) 11/12 divides by 2 5/8

11/12 divides by 21/8

11/12 * 8/21

22/63

13) 1 9/13 divides by 3/8

22/13 divides by 3/8

22/13 * 8/3

176/39

4 20/39

14) 6 4/5 divides by 3 2/9

34/5 divides by 29/2

34/5 * 2/29

68/145

15) 9 2/3 divides by 6 8/9

29/3 divides by 62/9

87/62

1 25/69

<h3>Division problem</h3>

Width of patio = 5 2/3 feet

Area of patio = 36 5/6 sq. feet

Length will be: area /width

36 5/6 divides by 5 2/3

221/6 divides by 17/3

221/6 * 3/17

13/2

6 1/2 feet

Length is 6 1/2 feet, that means we can place the concrete patio on a 7 foot picnic table.

17) Area of mirror = 225 sq. inches

Width of mirror = 13 3/4 inches.

Length of mirror will be=area/ width

225 divides by 13 3/4

225 divides by 55/4

225 * 4/55

180/11

16 4/11

The width of mirror will be 16 4/11 feet. It is greater than 16 feet. Therefore mirror can't be fit in a space of 15 inches by 16 inches.

18) Total length of fabric = 16 1/5 yards

One fabric piece size= 5 2/5

Number of fabric pieces will be= total length/ 1 fabric length

16 1/5 divides by 5 2/5

81/5 divides by 27/5

81/5*5/27

3

There will be total 3 pieces of size 5 2/5 yards.

3 0

3 years ago

Convert the following in standard form : -4x=4y-52

goldenfox [79]

The standard form of any linear equation with two variables is $ax+by=c$ , where a, b, and c are constants. To convert this equation to standard form, simply subtract $4y$ from both sides to get

$-4x-4y=-52$

We could leave it in this form, though, in this case, it might be better at if we multiplied both sides by -1 to get

$4x+4y=52$

8 0

2 years ago

Please help meee worth 10 pts

maxonik [38]

The answer is:
3. C
4. A
Hope this helps!

5 0

3 years ago

Read 2 more answers

Explain how multiplying with 6 is like multiplying with 3.

Debora [2.8K]

Multiplying with 6 is like multiplying with 3 because, you just double it. Let's say 3 × 4 = 12
Its the same as 6 × 2 = 12 ... which is just 3 + 3 + 3 + 3 . ((:

5 0

2 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

2 years ago