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Rzqust [24]
3 years ago
6

Use an array and partial products to find

Mathematics
1 answer:
creativ13 [48]3 years ago
3 0
Well done mate you’re correct.
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For which sample size (n) and sample proportion (p) can a normal curve be
Nonamiya [84]

Answer:

Option c.

Step-by-step explanation:

Using the normal curve to approximate a sampling distribution:

For a sample size n and a proportion n, the normal curve can be used if:

np \geq 10 and n(1-p) \geq 10

Option a:

np = 35*0.8 = 28 > 10

n(1-p) = 35*0.2 = 7 < 10

So option a cannot be used.

Option b:

np = 65*0.9 = 58.5 > 10

n(1-p) = 65*0.1 = 6.5 < 10

So option b cannot be used.

Option c:

np = 65*0.8 = 52 > 10

n(1-p) = 65*0.2 = 13 > 10

So option c can be used, and is the answer

Option d:

np = 35*0.9 = 31.5 > 10

n(1-p) = 35*0.1 = 3.5 < 10

So option d cannot be used.

The answer is given by option c.

7 0
3 years ago
Please!!!! HELP ME!! I give brainlyest!!! are u my friend? The table below models the cost, y, of using a high-efficiency washin
horsena [70]

Answer:

y= 25x+500

y=30x+400

20

standard machine

Step-by-step explanation:

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5 0
3 years ago
Which transformation could NOT be used to prove that two circles are congruent to one another?
EleoNora [17]

Answer: D

Step-by-step explanation:

6 0
3 years ago
What is the definition of a ray
Elan Coil [88]
Mathematically, a ray is a portion of a line which starts at a point and goes off in a particular direction to infinity.


4 0
4 years ago
There are 1,657 souvenir paperweights that need to be packed in boxes. Each box will hold 17
blondinia [14]

Answer:

Therefore we can say that 98 boxes will be needed to hold all the paperweights.

Step-by-step explanation:

i) there are 1657 souvenir paperweights that need to be packed in boxes.

ii) each box will hold 17 paperweights

iii) therefore the number of boxes that will be needed are

   = \dfrac{total\hspace{0.15cm} number\hspace{0.15cm} souvenir\hspace{0.15cm} paperweights}{number\hspace{0.15cm} of\hspace{0.15cm} paperweights\hspace{0.15cm} per\hspace{0.15cm} box}  = \dfrac{1657}{17} = 97.471

iv) Therefore we can say that 98 boxes will be needed to hold all the paperweights.

4 0
4 years ago
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