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andreev551 [17]

2 years ago

15

Which expression is equivalent to StartRoot StartFraction 128 x Superscript 5 Baseline y Superscript 6 Baseline Over 2 x Supersc

ript 7 Baseline y Superscript 5 Baseline EndFraction EndRoot?
Assume x Greater-than 0 and y > 0.
A.StartFraction x StartRoot y EndRoot Over 8 EndFraction
B.StartFraction y StartRoot x EndRoot Over 8 EndFraction
C.StartFraction 8 StartRoot x EndRoot Over y EndFraction
D.StartFraction 8 StartRoot y EndRoot Over x EndFraction

2 answers:

alexgriva [62]2 years ago

4 0

Answer:

C

Step-by-step explanation:

Anuta_ua [19.1K]2 years ago

4 0

Answer:

C

Step-by-step explanation:

You might be interested in

Joe scored 10 points in the basketball game. This is 2 more than one-sixth of the points scored by the team. Which equation, whe

marissa [1.9K]

(1/6) x + 2 = 10

Step-by-step explanation:

Step 1 :

Let x be the number of points scored by the team.

Given Joe has scored 2 points more than one sixth of the point scored bynthe team

Step 2:

Based on the above given information the equation which gives the points scored by Joe is as follows:

(1/6) x + 2

Given that Joe has scored 10 points, we have

(1/6)x + 2 = 10

Step 3:

When we solve the above equation for x , we get the total number of points scored by the team

7 0

2 years ago

Read 2 more answers

Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec

jeka94

Answer:

a) Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

Defect >2 = cannot be sold

Y = the number of defects on a fat quarter that can be sold by Company F.

Y = defect that can be sold

Y = Defect less or equal to 2 = 0,1,2

Probability distribution of the random variable Y:

Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean of Y (μ)

μ = Σ x*P(Y)

= (0*0.58) +(1*0.23)+(2*0.11)

= 0+0.23+0.22 = 0.45

Standard deviation of Y = σ

σ = Σ√(x-mean)^2*P(Y)

= Σ√[(x- μ )^2*P(Y)]

= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]

= √[0.11745 + 0.069575 +0.264275

= √(0.4513

σ = 0.6718

Company G:

σ for defect that be sold = 0.66

μ for defect that be sold = 0.40

Difference between μ of F and μ of G

= 0.45-0.40 = 0.05

Difference between σ of F and σ of G

= 0.67-0.66 = 0.01

Selling price of fat quarter without defect = $5

Discount per defect = $1.5

Selling price per defect = 5-1.5 = $3.5

Discount per 2 defect = $1.5*2 = $3

Selling price per defect = 5-3 = $2

Since defect to be sold cannot be greater than 2, let Y = 5,3,2

Probability distribution of the selling price Y:

Y 5 3 2

P(Y) 0.58 0.23 0.11

μ = (5*0.58) +(3.5*0.23)+(2*0.11)

μ = 2.9+0.805+0.22 =1.285

σ = Σ√[(x- μ )^2*P(Y)]

σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]

σ = 8.00+0.68+0.06 = 8.74

7 0

2 years ago

Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷

tester [92]

Answer: a) 278382 oz

b) 43 oz

c) 64.1 oz

Step-by-step explanation:

a) 8 T. 1,398 lb. 14 oz. × 6

First we need to change it in 'oz'.

As we know that

$1\ ton=32000\ oz\\\\1\ lb=16\ oz$

so, it becomes,

$8\times 32000+1398\times 16+14\ oz\\\\=278382\ oz$

8 T. 1,398 lb. 14 oz. × 6 becomes

$278382\times 6\\\\=1670292\ oz$

b) 349 lb. 6 oz. ÷ 130

It becomes,

$349\times 16+6\ oz\\\\=5590\ oz\\\\\text{ at last it becomes}\\\\=\frac{5590}{130}\\\\=43\ oz$

c) 6 T. 294 lb. ÷ 3,071

First it becomes,

$6\times 32000+294\times 16\ oz\\\\=196704\ oz$

At last it becomes,

$\frac{196704}{3071}\\\\=64.1\ oz$

Hence, a) 278382 oz

b) 43 oz

c) 64.1 oz

6 0

2 years ago

Read 2 more answers

This table represents a proportional relationship. Use the table to identify the constant of proportionality between the ratio o

VARVARA [1.3K]

Answer:

It would be 12...

Step-by-step explanation:

6 0

2 years ago

Liono4ka [1.6K]

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

$(x,y)\rightarrow (-x,y)$

It means the polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2:

If polygon MNOPQ is translated 3 units right and 5 units down, then

$(x,y)\rightarrow (x+3,y-5)$

$M(-2,8)\rightarrow V(-2+3,8-5)=V(1,3)$

$N(-4,12)\rightarrow W(-4+3,12-5)=W(-1,7)$

$O(-10,12)\rightarrow X(-10+3,12-5)=X(-7,7)$

$P(-8,8)\rightarrow Y(-8+3,8-5)=Y(-5,3)$

$Q(-6,6)\rightarrow Z(-6+3,6-5)=Z(-3,1)$

Therefore the vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

6 0

2 years ago