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Paha777 [63]
2 years ago
8

To approximate the number of bees in a hive, multiply the number of bees that leave the hive in one minute by 3 and divide by 0.

014. You count 25 bees leaving a hive in one minute. How many bees are in the hive?
There are about bees in the hive.
Mathematics
2 answers:
trasher [3.6K]2 years ago
8 0

Answer:

Approximatley 5,357 bees are in the hive

Step-by-step explanation:

AlekseyPX2 years ago
6 0

Answer:

There are about 5,357 bees in the hive

Step-by-step explanation:

Let

x ----->  the number of bees that leave the hive in one minute

y -----> the approximate number of bees in a hive

we know that

The formula to calculate the approximate number of bees in a hive is equal to

y=\frac{3x}{0.014}

For x=25

substitute

y=\frac{3(25)}{0.014}

y=5,357\ bees

therefore

There are about 5,357 bees in the hive

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mr Goodwill [35]

Answer:

53437.3639

Step-by-step explanation:

23y5o9-023.36839637836386378673676763767678362

6 0
2 years ago
How is the graph of y = negative RootIndex 3 StartRoot x minus 4 EndRoot transformed to produce the graph of y = negative RootIn
SVEN [57.7K]

Answer:

Multiply by ∛2 and translate the graph to left by 4 units.

Step-by-step explanation:

The initial function given is:

y = -∛(x - 4)

The transformed function is:

y = -∛(2x - 4)

Consider the initial function.

y = -∛(x - 4)

(Represented by Black line in the graph)

Multiply the function by ∛2. The function becomes:

y = -∛(x - 4) × ∛2

y = -∛(2)(x-4)

y = -∛(2x-8)

(Represented by Red line in the graph represents this function)

Translate the graph 4 units to the left by adding 4 to the x component:

y = -∛(2x-8+4)

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(Represented by Blue line in the graph)

3 0
2 years ago
Read 2 more answers
A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

7 0
2 years ago
The owner of a football team claims that the average attendance at games is over 74,900, and he is therefore justified in moving
Slav-nsk [51]

Answer:

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Step-by-step explanation:

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= 74,900

We will have to test:

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or,

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Verses,

⇒ H_1: \mu> \mu_0

or,

    H_1: \mu >74,900

The other given alternatives aren't connected to the given scenario. So the above is the correct one.

4 0
2 years ago
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2 years ago
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