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Lostsunrise [7]

3 years ago

9

the entry ticket for a baseball game costs $5 for an adult ticket and $3 for a child.A group of 20 people paid total entry fees

of $90

2 answers:

sesenic [268]3 years ago

7 0

There would be 15 adults and 5 children in that group.
15 adults = $75
5 Children= $15
75+15= $90

Andreas93 [3]3 years ago

6 0

Sooo 5 children and 15adults

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The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6

viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The proportion of contracts involved between five and 10 inclusive bidders is 0.5.

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders No. contracts Relative frequency of contracts

2 6 6/102=0.0588

3 20 20/102=0.1961

4 24 24/102=0.2353

5 18 18/102=0.1765

6 13 13/102=0.1275

7 7 7/102=0.0686

8 5 5/102=0.049

9 6 6/102=0.0588

10 2 2/102=0.0196

11 1 1/102=0.0098

Total 102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The proportion of contracts involved between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10= Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

8 0

3 years ago

What is the sum or difference?

dlinn [17]

1. Ans: Option (B) $-6x^4$

Explanation:
Given: $2x^4 - 8x^4$
Take the common out:
=> $x^4(2 - 8)$

Hence: => $[tex]-6x^4$ [/tex] (Option B)

2. Ans: Option (D) $-3y^5$

Explanation:
Given: $6y^5 - 9y^5$
Take the common term(s) out:
=> $y^5(6-9)$

Hence: => $-3y^5$ (Option D)

3. Ans: Option (B) $9x^2 - 12x + 6$ <span> quadratic trinomial

Explanation:
</span>Given: $6 - 12x + 13x^2 - 4x^2$

The standard form of polynomial function must have the highest powered value at the start, then the second highest and so on.

=> $13x^2 - 4x^2 - 12x + 6$
=> $9x^2 - 12x + 6$ (Option B)

4. Ans: Option (C) $7.3x^2 + 0.8x + 1.3$

Explanation:
In order to find the total number of C<span>ommon and Endler's guppies, you need to add both Common and Endler's guppies' polynomials, as follows:

</span>Common guppies: $3.1x^2 +6.0x + 0.3$ <span>
Endler's guppies: </span> $4.2x^2 - 5.2x + 1.0$
(add both)
-----------------------------------------------------------------
Total number: $7.3x^2 +0.8x + 1.3$
-----------------------------------------------------------------
Hence the ans is Option(C) $7.3x^2 + 0.8x + 1.3$

5. Ans: Option (D) $2x^2(133 - 18 \pi )$

Explanation:
First let's find the total area of the yard:
Total Area of the Yard = 14x * 19x = $266x^2$

Now the area of the circular fountain:
Area of the Circular Fountain = $\pi r^2$
Since, r=6x
Therefore,
Area of the Circular Fountain = $\pi (6x)^2 = 36 \pi x^2$

Now the final Area of the yard would be:
Final area of the Yard = Total Area of the Yard - Area of the Circular Fountain
Final area of the Yard = $266x^2$ - $36 \pi x^2$
=> Final area of the Yard = $2x^2(133 - 18 \pi)$ (Option D)

6. Ans: Option (A) $56x^2$

Explanation:
First let's find the total area of the lot:
Total Area of the lot= 6x * 10x = $60x^2$

Now the area of the stadium:
Area of the stadium = 1x * 4x = $4x^2$

Now the final Area of the lot would be:
Final area of the lot= Total Area of the lot - Area of the stadium
Final area of the lot= $60x^2$ - $4x^2$
=> Final area of the lot = $56x^2$ (Option A)

-i

4 0

3 years ago

Read 2 more answers

No links, ty <3

levacccp [35]

Answer:

sAy yOu aRe mY bAkA

Step-by-step explanation:

ty for the points <3

4 0

3 years ago

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.Refer to

Dafna11 [192]

Given that

the weight of football players is distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

And we need to find What is the minimum weight of the middle 95% of the players?

Explanation -

Using the Empirical Rule, 95% of the distribution will fall within 2 times of the standard deviation from the mean.

Two standard deviations = 2 x 25 pounds = 50 pounds

So the minimum weight = 200 pounds - 50 pounds = 150 pounds

Hence the final answer is 150 pounds.

3 0

1 year ago

When renting a limo for prom, the number of people varies inversely with the cost per person. Originally there were 12 people an

pashok25 [27]

The answer would be $10.

8 0

3 years ago