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Furkat
1 month ago
12

What proportion of these candy bars have more than 300 calories per serving? (Round your answer to two decimal places.)

Mathematics
1 answer:
Inessa [12.5K]1 month ago
3 0

Answer: 0.12

Step-by-step explanation:

There are a total of 65 candy bars. Within this amount;

2 candy bars contain 300 to 350 calories

1 candy bar contains 350 to 400 calories

4 candy bars contain 400 to 450 calories

1 candy bar contains 450 to 500 calories

Thus, the overall ratio of candy bars with more than 300 calories is;

= (2 + 1 + 4 + 1) / 65

= 8/65

= 0.12

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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
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Answer:

Step-by-step reasoning:

(a)

To have an accepted bid, it needs to exceed $10,000. Let bid x be a continuous random variable uniformly distributed between

$10,000 and $15,000

The range for accepted bidding is [ {\rm{\$ 10,000, \$ 15,000}], with b = $15000 and a = $10000.

The provided bidding range is [$10,000,$12,000]. The probability is determined as follows,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The accepted bidding range is [$10,000,$15,000], where b = $15,000 and a =$10,000. The given bidding range is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The optimal amount to bid for maximizing the probability of acquiring the property is calculated as,  

The accepted bidding range is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The provided bidding range is [$10,000,$15,000].

\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}

(d)  If you know someone willing to pay you $16,000 for the property, would you still consider bidding less than the amount mentioned in part (c)? Why or why not?

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1 month ago
A comic book store is having a sale you can buy 20 comic books for $35. what is the cost of 8 comic books during the sale
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Response:

$14

Step-by-step explanation:

35 divided by 20 equals 1.75

8 multiplied by 1.75 results in 14

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27 days ago
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Morgan, Dakota and Taylor showed their work solving the equation 7x + 5x +1= 73. Identify the student with the invalid first ste
tester [12383]

Answer:

Morgan, Dakota, and Taylor demonstrated their calculation for the problem 7x + 5x + 1 = 73. Point out which student made an incorrect initial step and explain your reasoning. Morgan: 7x + 5x + 1 = 73 12x + 1 = 73 Dakota: 7x + 5x + 1 = 73 7x + 6x = 73 Taylor: 7x + 5x + 1 = 73 7x + 5x = 72

Step-by-step explanation:

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12 days ago
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Apply the distributive property to factor out the greatest common factor. 32 + 44 = 32+44=32, plus, 44, equals
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For your college interview, you must wear a tie. You own 3 regular (boring) ties and 5 (cool) bow ties. 36 1. Counting (a) How m
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Response:

a) 8

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Detailed explanation:

Section a) Options for neck-wear

Regular ties count = 3

Bow ties count = 5

Overall ties count = 3 + 5 = 8

Any tie can be selected from these 8 available options. Thus, the options for neck-wear total 8.

Section b)

We now calculate choices for using both regular and bow ties.

Selecting a regular tie is independent of the bow tie selection. According to counting principles, if two events are independent, the total ways to realize both equal the product of their individual possibilities. Thus,[

Number of ways to select both ties = Count of ways to choose each tie

So,

Ways for wearing both types of ties = 3 x 5 = 15 ways

Section c)

Count of shirts = 5

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Count of pants = 3

Count of dresses = 7

Choice options for outfits include:

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Pants can be chosen in 3 ways. Pairing a skirt or pant provides 9 options.

Reapplying counting rules:

Ways to wear a shirt with a skirt or pant = 3 x 9 = 27

Choices for dresses = 7

Consequently, the total is 27 + 7 = 34 outfit combinations.

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19 days ago
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