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riadik2000 [5.3K]
3 years ago
8

A candy company produces individually wrapped candies. The quality control manager for the company believes that the weight of t

he candies is approximately normally distributed with mean 720 milligrams (mg).If the manager’s belief is correct, which of the following intervals of weights will contain the largest proportion of the candies in the distribution of weights?
a) 740 mg to 780 mgb) 700 mg to 740 mgc) 680 mg to 720 mgd) 660 mg to 720 mge) 620 mg to 660 mg
Mathematics
1 answer:
Rama09 [41]3 years ago
6 0

Answer:

b) 700 mg to 740 mg

Step-by-step explanation:

The mean is the sum of all the samples divided by the quantity of the samples, so a mean is a number between the maximum and the minimum sample. In this case, 720 mg must be in the interval, so letter<em> a</em> and<em> e </em>are eliminated.

The mean also represents the proportion of the samples, because the samples intend to be close to the mean. So, if 720 mg is the maximum or the minimum sample weight, it means that there are many samples close to it, and the proportion of the distribution will be small. So, letter <em>c</em> and <em>d</em> can't be the answer.

So, the largest proportion of weights is given in letter b, the interval of weights must be 700 mg to 740 mg.

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The circumference of Circle K is $\pi$ . The circumference of Circle L is $4\pi$ . Two circles, one labeled "Circle K" and the o
timurjin [86]

Answer:

Step-by-step explanation:

Given the circumference of Circle K = π

circumference of Circle L = 4π

Ratio of their circumferences = Ck/Cl

Ratio of their circumferences = π/4π

Ratio of their circumferences = 1/4 = 1:4

For their radii

C = 2πr

for circle k with circumference π

π = 2πrk

1 = 2rk

rk = 1/2

for circle l with circumference 4π

4π = 2πr

4 = 2r

r = 4/2

rl = 2

ratio

rk/rl = 1/2/2

rk/rl = 1/4 = 1:4

for the areas

Area of a circle = πr²

for circle k

Ak = π(1/2)²

Ak = π(1/4)

Ak = π/4

for circle l

Al = π(2)²

Al = 4π

Ratio of their areas

Ak/Al = π/4/(4π)

Ak/Al = π/16π

Ak/Al = 1/16 = 1:16

3 0
2 years ago
Write a polynomial function, p(x) with degree 3 that has p(7)=0
MArishka [77]

Answer:

p (x) = x^{3} - 21x^{2}+ 147x - 343

is the required polynomial with degree 3 and p ( 7 ) = 0

Step-by-step explanation:

Given:

p ( 7 ) = 0

To Find:

p ( x ) = ?

Solution:

Given p ( 7 ) = 0 that means substituting 7 in the polynomial function will get the value of the polynomial as 0.

Therefore zero's of the polynomial is seven i.e 7

Degree : Highest raise to power in the polynomial is the degree of the polynomial

We have the identity,

(a -b)^{3} = a^{3}-3a^{2}b +3ab^{2} - b^{3}

Take a = x

        b = 7

Substitute in the identity we get

(x -7)^{3} = x^{3}-3x^{2}(7) +3x(7)^{2} - 7^{3}\\(x -7)^{3} = x^{3}-21x^{2} +147x - 343

Which is the required Polynomial function in degree 3 and if we substitute 7 in the polynomial function will get the value of the polynomial function zero.

p ( 7 ) = 7³ - 21×7² + 147×7 - 7³

p ( 7 ) = 0

p (x) = x^{3} - 21x^{2}+ 147x - 343

4 0
3 years ago
Pls help me I don’t understand this
I am Lyosha [343]
You didn’t put a question
3 0
2 years ago
Read 2 more answers
Find cscθ.<br><br>pls answer
ExtremeBDS [4]

Answer:

The answer to your question is csc Ф = \frac{-5}{4}

Step-by-step explanation:

Process

1.- Determine the sign

We must determine csc Ф in the forth quadrangle, here csc is negative.

2.- Determine the hypotenuse

     c² = a² + b²

     c² = 3² + (-4)²

     c² = 9 + 16

     c² = 25

     c = 5

3.- Determine csc Ф

    csc Ф = \frac{hypotenuse}{opposite side}

    csc Ф = \frac{5}{-4}  = \frac{-5}{4}

7 0
2 years ago
A home with a market value of $240,000 is assessed at 40% of the market value. What's the assessed value?
hammer [34]

Answer:

  $96,000

Step-by-step explanation:

40% of $240,000 is ...

  0.40 × $240,000 = $96,000

The assessed value is $96,000.

4 0
3 years ago
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