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2 years ago

7

The following table shows how many pies 444 baking teams made at a pie-baking competition. For example, Team Crusty baked 282828

pies in 888 hours.

2 answers:

lys-0071 [83]2 years ago

6 0

The answer would be B. I divided the pies per hour

Or : pies / hour = ??

For the “pie piece” team, 27 / 6 = 4.5

Brilliant_brown [7]2 years ago

3 0

Answer:

The answer would be B

Step-by-step explanation:

Divide each number of pies made by each team by 4.5

Which one has a quotient of the same number of hours they were baking pies is the correct answer.

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One week a computer store sold a total of 36 computers and external hard drives. the revenue from these sales was $30875 . if c

melamori03 [73]

To solve this problem you must apply the proccedure shown below:

1. You must make a system of equation by using the information given in the problem above. Then, let's call:

$x$ : The number of computers sold.

$y$ : The number of hard drives sold.

2. Therefore, you have the following system of equations:

$\left \{ {{x+y=36} \atop {1180x+125y=30875}} \right.$

2. Solve for $x$ in the first equation and substitute it into the second equation:

$x=36-y\\ 1180(36-y)+125y=30875$

3. Solve for $y$ :

$42480-1180y+125y=30875\\ y=11$

4. Substitute the value of $y$ into the first equation to calculate $x$ :

$x+11=36\\ x=25$

Therefore, the answer is: The store sold $25$ computers and $11$ hard drives.

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3 years ago

Use the drop-down menus to describe the key aspects of the function f(x) = –x2 – 2x – 1.

Aliun [14]

Answer:

vertex is the (maximum value)

function is increasing (when x<-1)

function is decresing (when x>-1)

domain of the function is (all real numbers)

range of the function is (all numbers less than or equal to 0)

Step-by-step explanation:

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2 years ago

How do you solve an equation with 2 variables on one side and and the answer on the other

Irina-Kira [14]

Answer: combine like terms dude

Step-by-step explanation:

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3 years ago

A researcher randomly assigned N = 16 rodents to experience one of four levels of shock (n = 4 per group) following the illumina

Umnica [9.8K]

Answer:

The hypothesis should be retained

Step-by-step explanation:

We are given:

N= 16

n = 4

$SS_B_G = 24$

$SS_E = 48$

Level of significance, a = 0.05

From the given data, our ANOVA table will be:

$----Source: Between---- Residual---$

$------DF:n-1=> 4-1=3------N-n=>16-4=12$

$------SS: SS_B_G = 24---- SS_E =48$

$----MS: MS_B_G = 8----MS_E=48/12=4$

Note: proper ANOVA table is attached

The statistic will be written as:

$TS = \frac{MS_B_G}{MS_E} ~ F_n_-_1, _N_-_n$

T = 8/4 = 2

Therefore

$F_n_-_1, _N_-_n, _a$

$=> F_3,_1_2,_ 0_._0_0_5 = 3.490$

The hypothesis should be retained, because we no know that the statistic is in the acceptance area.

7 0

3 years ago

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casi

boyakko [2]

The missing part in the question;

and the payoff is $2.20 won for every dollar bet. (As the player’s probability of winning in this case is $\dfrac{1}{4}$ ........

Also:

For such a bet, the casino pays off as shown in the following table.

The table can be shown as:

Keno Payoffs in 10 Number bets

Number of matches Dollars won for each $1 bet

0 - 4 -1

5 1

6 17

7 179

8 1299

9 2599

10 24999

Answer:

Step-by-step explanation:

Given that:

Twenty numbers are selected at random by the casino from the set of numbers 1 through 80

A player can select from 1 to 15 numbers; a win occurs if some fraction of the player’s chosen subset matches any of the 20 numbers drawn by the house

Let assume X to represent the numbers of player chooses which are in the Casino-selected-set of 20.

Let assume the random variable X has a hypergeometric distribution with parameters N= 80 and m =20.

Then, the probability mass function of a hypergeometric distribution can be defined as:

$P(X=k)=\dfrac{(^m_k)(^{N-m}_{n-k})}{(^N_n)}, k =1,2,3 ... n$

Now; the probability that i out of n numbers chosen by the player among 20 can be expressed as:

$P(X=k)=\dfrac{(^{20}_k)(^{60}_{n-k})}{(^{80}_n)}, k =1,2,3 ... n$

Also; given that ; When the player selects 2 numbers, a payoff (of odds) of $12 won for every $1 bet is made when both numbers are among the 20

So; n= 2; k= 2

Then :

Probability P ( Both number in the set 20) $=\dfrac{(^{20}_2)(^{60}_{2-2})}{(^{80}_2)}$

Probability P ( Both number in the set 20) $= \dfrac{20*19}{80*79}$

Probability P ( Both number in the set 20) $=\dfrac{19}{316}$

Probability P ( Both number in the set 20) $=\dfrac{1}{16.63}$

Thus; the payoff odd for $=\dfrac{1}{16.63}$ is 16.63:1 ,as such fair payoff in this case is $16.63

Again;

Let assume X to represent the numbers of player chooses which are in the Casino-selected-set of 20.

Let assume the random variable X has a hypergeometric distribution with parameters N= 80 and m =20.

The probability mass function of the hypergeometric distribution can be defined as :

$P(X=k)=\dfrac{(^m_k)(^{N-m}_{n-k})}{(^N_n)}, k =1,2,3 ... n$

Now; the probability that i out of n numbers chosen by the player among 20 can be expressed as:

$P(n,k)=\dfrac{(^{20}_k)(^{60}_{n-k})}{(^{80}_n)}, k =1,2,3 ... n$

From the table able ; the expected payoff can be computed as shown in the attached diagram below. Thanks.

8 0

3 years ago