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Olin [163]
3 years ago
12

if an orange gets cut in half how much do you have left if you cut in forths how many slices are there​

Mathematics
2 answers:
sammy [17]3 years ago
6 0

Answer:

8 slices

Step-by-step explanation:

a_sh-v [17]3 years ago
4 0

Answer:

There are 10 slices left

Step-by-step explanation:

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4a x 6a⁵ ÷ 8a²<br><br>please answer the question in full steps, thank you ​
german

Answer:

4a×6a^5 ÷8a^2

24a^6÷8a^2

3a^4

6 0
3 years ago
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Find f(3).<br><br> f(x) = x2 + 5x - 9<br> A) 7 <br> B) 10 <br> C) 9 <br> D) 15
Lelechka [254]

Answer:

f(3) = 15. Correct: D)

Explanation:

<u>Numeric Value of a Function</u>

The value of a function f(x) when x = a is calculated by replacing the x for a. We have the function:

f(x)=x^2+5x-9

It is required to find f(3), or the numeric value of f when x=3. Replace x for 3

f(3)=3^2+5(3)-9

f(3)=9+15-9

f(3)=15

A) Incorrect. f(3) is not 7

B) Incorrect. f(3) is not 10

C) Incorrect. f(3) is not 9

D) Correct. f(3) =15 as found above.

4 0
3 years ago
Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. ​(a) Descr
Daniel [21]

Answer:

(a) The sampling distribution of\overline{X} = Population mean = 79

(b)  P ( \overline{X} greater than 81.2 ) =  0.1357

(c) P (\overline{X} less than or equals 74.4 ) = .0107

(d) P (77.6 less than \overline{X} less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean (\nu) = 79

Standard deviation (\sigma ) = 18

​(a) Describe the sampling distribution of \overline{X}

For large sample using central limit theorem

the sampling distribution of\overline{X} = Population mean = 79

​(b) What is Upper P ( \overline{X} greater than 81.2 )​ =

P(\overline{X}> 81.2)  = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})

                    =  P(Z> 1.1)

                    = 1 - P(Z<   1.1)

                    = 1 - .8643 =

                    = 0.1357

(c) What is Upper P (\overline{X} less than or equals 74.4 ) =

P(\overline{X}\leq  74.4) = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{74.4- 79}{\frac{18}{\sqrt{81}}})

                    = P(Z\leq  -2.3)

                    = .0107

​(d) What is Upper P (77.6 less than \overline{X} less than 83.2 ) =

P(77.6< \overline{X}<   83.2) = P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{83.2- 79}{\frac{18}{\sqrt{81}}})

                                = P(- 0.7< Z<   2.1)

                                 = (Z<   2.1) - (Z<   -0.7)

                                  = 0.9821 - .2420

                                   = 0.7401

3 0
3 years ago
Help please i need to know this
HACTEHA [7]

Answer:true,true,false, true,false false

Step-by-step explanation:

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Your number would be 10 and the exponet would be 6
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