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Pachacha [2.7K]
3 years ago
5

I just need to know the answer

Mathematics
1 answer:
Alex777 [14]3 years ago
3 0
To which one?…………………………… what number?
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Which of the following is the solution to 17 - 2x = -11?
Yuliya22 [10]

Answer:

(D)14

Step-by-step explanation:

17-2x=-11

17+11=2x

28=2x

28/2=2x/2

14=x

x=14

8 0
3 years ago
Read 2 more answers
Which number is equivalent to 17.02?<br> 17.020 , 17.22 , 17 , 17.002
Irina18 [472]

Answer:

Step-by-step explanation:

17.020

3 0
3 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
Mr. Williams is building a sand box for his children. It costs $228 for the sand if he builds a rectangular-sand box with dimens
Andru [333]

Answer:  The correct option is (C) $342.

Step-by-step explanation:  Given that Mr. Williams is building a sand box for his children and is costs $228 for the sand if he builds a rectangular-sand box with dimensions 9 ft by 6 ft.

We are to find the cost of the sand if he decides to increase the size to 13\frac{1}{2}~\textup{ft by }9~\textup{ft}.

Since the box is empty from inside, so we will be considering the perimeter of the box, not area.

The perimeter of the rectangular-sand box with dimensions 9 ft by 6 ft is

P_1=2(9+6)=30~\textup{ft},

and the perimeter of the rectangular-sand box with dimensions 13\frac{1}{2}~\textup{ft by }9~\textup{ft}. is

P_2=2\left(13\dfrac{1}{2}\times9\right)=2(13.5\times9)=45~\textup{ft}.

Now, we will be using the UNITARY method.

Cost of sand for building rectangular-sand box with perimeter 30ft = $228.

So, cost of  sand for building rectangular-sand box with perimeter 1 ft will be

\$\dfrac{228}{30}.

Therefore, the cost of sand for building rectangular-sand box with perimeter 45 ft is given by

\$\dfrac{228}{30}\times45=\$342.

Thus, the required cost of the sand is $342.

Option (C) is CORRECT.

3 0
3 years ago
Read 2 more answers
If the radius of a circle is 5 what is the circumference
Pachacha [2.7K]

Answer:

31.42

hope this helps

have a good day :)

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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