The three perpendicular bisectors of the sides of AABC intersect at point G. Point G is the

Fifty-four is ___% of 60. 20 points offered

prisoha [69]

Answer:

54÷60×100= 90%

Step-by-step explanation:

90%

cmon man

4 0

4 years ago

Read 2 more answers

Find the product. (-3ab2)3 -27 a3b6 -9 a3b5 -23 a2b2 27 ab

7nadin3 [17]

Answer:

a) -27 a³ b⁶

Step-by-step explanation:

Explanation:-

Given ( -3 ab² )³

By using (ab )ⁿ = aⁿ bⁿ

( -3 ab² )³ = (-3)³ a³ (b²)³

Again , using formula $(a^{m} )^{n} = a^{mn}$

= -27 a³ b⁶

6 0

3 years ago

Cube root of -6×-6 to the power of 2+4 times -6

Orlov [11]

This is what my calculator says

8 0

4 years ago

Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams

Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0

3 years ago

Which score indicates the highest relative position? I. A score of 2.6 on a test with X = 5.0 and s = 1.6 II. A score of 650 on

Zanzabum

Answer:

A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6 and A score of 48 on a test with $\bar X$ = 57 and s = 6 indicate the highest relative position.

Step-by-step explanation:

We are given the following:

I. A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6

II. A score of 650 on a test with $\bar X$ = 800 and s = 200

III. A score of 48 on a test with $\bar X$ = 57 and s = 6

And we have to find that which score indicates the highest relative position.

For finding in which score indicates the highest relative position, we will find the z score for each of the score on a test because the higher the z score, it indicates the highest relative position.

The z-score probability distribution is given by;

Z = $\frac{X-\bar X}{s}$ ~ N(0,1)

where, $\bar X$ = mean score

s = standard deviation

X = each score on a test

The z-score of First condition is calculated as;

Since we are given that a score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6,

So, z-score = $\frac{2.6-5}{1.6}$ = -1.5 {where $\bar X = 5.0$ and s = 1.6 }

The z-score of Second condition is calculated as;

Since we are given that a score of 650 on a test with $\bar X$ = 800 and s = 200,

So, z-score = $\frac{650-800}{200}$ = -0.75 {where $\bar X = 800$ and s = 200 }

The z-score of Third condition is calculated as;

Since we are given that a score of 48 on a test with $\bar X$ = 57 and s = 6,

So, z-score = $\frac{48-57}{6}$ = -1.5 {where $\bar X = 57$ and s = 6 }

AS we can clearly see that the z score of First and third condition are equally likely higher as compared to Second condition so it can be stated that A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6 and A score of 48 on a test with $\bar X$ = 57 and s = 6 indicate the highest relative position.

7 0

4 years ago