Find the solution to the system of equations. ([?], [ ]

IN - Change this fraction to a decimal: 9

vazorg [7]

Answer: 9 over 10

Step-by-step explanation:

7 0

3 years ago

Joe nas to take two quizzes tomorrow. He nas only enough time to study one of them. • The science quiz has 7 true-or-false quest

Nostrana [21]

Answer:

C.

Step-by-step explanation:

Science questions are T/F. That's .5 or $\frac{1}{2}$ chance to get it right.

7 questions. Each with a $\frac{1}{2}$ .

That's $(\frac{1}{2})^{7}$

$(\frac{1}{2})^{7}$ = $\frac{1}{128}$ = .0078 = .008 rounded

English questions are multiple choice. 4 choices. That's .25 or $\frac{1}{4}$

3 questions. Each $\frac{1}{4}$

That's $(\frac{1}{4})^{3}$

$(\frac{1}{4})^{3}$ = $\frac{1}{64}$ = .0156 = .016 rounded

6 0

2 years ago

Can someone help please

marysya [2.9K]

I would say B. The mean is a good measure of central tendency.

I hope this helps!

6 0

3 years ago

Solve using the pathagorean theorem

seropon [69]

Answer:

2 sqrt(13)

Step-by-step explanation:

We can find the hypotenuse using the Pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

4^2 + 6^2 = c^2

16+36 = c^2

52 = c^2

Taking the square root of each side

sqrt (52) = c

sqrt(4*13) = c

2 sqrt(13) = c

4 0

2 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