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3 months ago
13

Complete the squares so that the differences shown on the outside are correct

Mathematics
2 answers:
zzz [12.3K]3 months ago
4 0

8 5 and 18 8

7 2 9 2

The differences calculated are 8 - 5 = 3, 8 - 7 = 1, 5 - 2 = 3, and for the other set, 18 - 8 = 10, 18 - 9 = 9, 9 - 2 = 7, 8 - 2 = 6


AnnZ [12.3K]3 months ago
3 0

Response:

8-1=7

8-3=5

7-6=1

10-9=1

Detailed explanation:

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Jacque needs to buy some pizzas for a party at her office. She's ordering from a restaurant that charges a \$7.50$7.50dollar sig
Svet_ta [12734]

Answer:

Step-by-step explanation:

Delivery charge = $7.50

Cost per pizza = $14

Budgeted total = $60

7.50 + 14p < 60

First, subtract 7.50 from both sides

14p < 60 - 7.50

14p < 52.50

Now divide both sides by 14

p < 52.50 / 14

p < 3.75

Each pizza consists of 8 slices

Therefore, the total number of slices she can buy = 3.75 × 8

= 30 slices

5 0
3 months ago
A scientist is studying some rabbits. A disease is killing the rabbits. A population of 240 of these rabbits was reduced to 180
Zina [12379]
By the seventh day, there are zero rabbits remaining.
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2 months ago
You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if
Zina [12379]

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)=\dfrac{1}{4}=0.25

You win $6 if both faces are heads, so P(Y=6)=P(HH)=\dfrac{1}{4}=0.25

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)=\dfrac{2}{4}=0.50

To find the expected value to win: E(Y)=\sum_{i=1}^{i=3} y_ip(y_1)

=1\times0.25+6\times0.25+(-3)\times0.50=0.25

Thus, the mean of Y: E(Y)= $0.25

E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25

Variance = E[Y^2]-E(Y)^2

=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19

Therefore, variance of Y = $ 6.19

6 0
2 months ago
Biologists stocked a lake with 240 fish and estimated the carrying capacity (the maximal population for the fish of that species
Inessa [12570]

Answer:

2.30 years

Step-by-step explanation:

The fish population increased threefold in the first year, resulting in 240 * 3 = 720 fish.

(a) The logistic equation can be represented as follows

P = P_0e^{kt}

where P0 = 240 denotes the initial fish count; substituting P = 720 and t = 1 allows us to determine the constant k

720 = 240e^{1k}

e^k = 3

k = ln3 = 1.1

b) By applying the formula below

P = P_0e^{kt}

with P set to 3000, P0 as 240, and k equal to 1.1, we can compute the duration needed for the population to reach 3000 fish

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1.1t = ln12.5 = 2.53

t = 2.30 years

6 0
2 months ago
Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.
babunello [11817]

Answer:

The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.

Step-by-step explanation:

Prior concepts

A normal distribution is characterized as a "probability distribution that is symmetric around the mean, indicating that data close to the mean are more frequent than those further away".

The Z-score refers to "a statistical measurement that reflects the relationship of a value to the mean of a group, measured in standard deviations".

Let X denote the random variable of interest, and we identify its distribution:

X \sim N(\mu=15,\sigma=4)

Also, let \bar X signify the sample mean, whose distribution is:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

In this case, \bar X \sim N(15,\frac{4}{\sqrt{64}})

Solution to the problem

We seek this probability

P(13.5

Applying the Z-score formula to the probability results in:

P(13.5

=P(\frac{13.5-15}{\frac{4}{\sqrt{64}}}

To determine these probabilities, we can refer to normal distribution tables, use Excel, or a calculator.

P(-3

The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.

7 0
2 months ago
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