What is √192 expressed in simplest radical form?

Jeksisososososodododo 5+2

Ivenika [448]

Answer: 7

Explanation:

5+1=6

6+1=7

1+1=2

5+2=7

7 0

3 years ago

Read 2 more answers

I need to finish this quick!!!!!!!!!!!!!!

ioda

Answer:

-4??

Step by step explanation:

i believe

6 0

3 years ago

The area of a square floor on a scale drawing is 36 square centimeters, and the scale drawing is

goldfiish [28.3K]

Area of a square = x² ( x is the length of a side)

Area = 36cm² = x² = 6²

length of one side of the square in the scale drawing = 6cm

since the diagram is scaled to 1cm : 3ft

the actual length = 6 × 3 ft = 18ft

the actual area = x² = 18² = 324 ft²

the ratio of the area in the drawing to the

actual area is

36 cm² : 324 ft²

to get the simplest ratio we divide both sides by 36 ( as 36 is the factor they both share in common)

the simplest ratio of the area in the drawing to the

actual area is

1 : 9

Hope you understood :)

6 0

2 years ago

Choose the correct option that explains what steps were followed to obtain the system of equations below.

Alika [10]

To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 5. The solution to system B will be the same as the solution to system A.

<h3>How to find the relationship between two equations?</h3>

System A equations:

-x - 2y = 7 ------ (1)

5x - 6y = -3 ------(2)

We multiply the first equation by 5, to get :

-5x - 10y = 35 -----(3)

The sum of equation (2) and equation (3) gives:

-16y = 32

y = 32/-16

y = -2

Thus, solution of system B is;

-x - 2(-2) = 7

-x + 4 = 7

x = -3

So, the solution of system B is (-3, -2), that means the solution of both systems are same.

Read more about equations at; brainly.com/question/2972832

#SPJ1

8 0

2 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of small cars that were totaled = $\frac{8}{12}$ = 0.67

$\hat p_2$ = sample proportion of large cars that were totaled = $\frac{5}{15}$ = 0.33

$n_1$ = sample of small cars = 12

$n_2$ = sample of large cars = 15

$p_1$ = population proportion of small cars that are totaled

$p_2$ = population proportion of large cars that were totaled

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

So, 95% confidence interval for the difference between population population, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < $p_1-p_2$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ , $(0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ ]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0

4 years ago