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2 years ago

PLEASE HELP NOW 20pts........

Mathematics

2 answers:

OlgaM077 [116]2 years ago

6 0

x = 4y – 2

1.5x + 5y = 1152

julia-pushkina [17]2 years ago

6 0

Answer:

number of tickets of $5 sold = 418 and number of tickets of $1.5 sold = 105

Step-by-step explanation:

Let y be the number of $5 tickets sold and x be the number of $1.50 tickets sold.

Now from the statement " Number of $1.50 tickets sold was 2 less than the four times the number of $5 tickets sold".

Equation will be x = 4y - 2

4y - x = 2 ---------(1)

For second equation statement is " The pep club raised $1152 from the ticket sales".

Equation will be 5y + (1.50)x = 1152 -------(2)

Now we multiply equation (1) by 1.5 and add it to equation (2)

6y - 1.5x + (5y + 1.5x) = 3 + 1152

11y = 1155

y = 105

Now we put y = 105 in equation (1)

x = 4×105 - 2

x = 420 -2

x = 418

Therefore, number of tickets of $5 sold = 418 and number of tickets of $1.5 sold = 105

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Len [333]

Answer:

(a) n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of n = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of n = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of n = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of n = 500

$P(\mu-200$

$=P(-1.86$

n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

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2 years ago

During a field trip, 60 students are put into equal-sized groups. Describe two ways to interpret 60÷5 60 ÷ 5 in this context. Fi

Afina-wow [57]

The second question:

Consider the division expression $7\frac{1}{2} / 2$ . Select all multiplication equations that correspond to this division expression.

$2 * ? = 7\frac{1}{2}$ $7\frac{1}{2} * ?= 2$ $? * 2 = 7\frac{1}{2}$

$2* 7\frac{1}{2} = ?$ $? * 7\frac{1}{2} = 2$

Answer:

1. See Explanation

2. $2 * ? = 7\frac{1}{2}$ and $? * 2 = 7\frac{1}{2}$

Step-by-step explanation:

Solving (a):

Given

$Students = 60$

$Group = Equal\ Sized$

Required

Interpret $\frac{60}{5}$ in 2 ways

Interpretation 1: Number of groups if there are 5 students in each

Interpretation 2: Number of students in each group if there are 5 groups

Solving the quotient

$Quotient = \frac{60}{5}$

$Quotient = 12$

For Interpretation 1:

The quotient means: 12 groups

For Interpretation 2:

The quotient means: 12 students

Solving (b):

Given

$7\frac{1}{2} / 2$

Required

Select all equivalent multiplication equations

Let ? be the quotient of t $7\frac{1}{2} / 2$

So, we have:

$7\frac{1}{2}/2 = ?$

Multiply through by 2

$2 * 7\frac{1}{2}/2 = ? * 2$

$7\frac{1}{2} = ? * 2$

Rewrite as:

$? * 2 = 7\frac{1}{2}$ --- This is 1 equivalent expression

Apply commutative law of addition:

$2 * ? = 7\frac{1}{2}$ --- This is another equivalent expression

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2 years ago

1. Dennis made a 500 km trip in five hours. For the first two hours, Dennis had an average speed of 150 km/h. For the last two h

Sati [7]

Answer:

The correct option is;

Dennis must have stopped for an hour in the middle of this trip

Step-by-step explanation:

The given parameters are;

The distance Dennis covered in 5 hours = 500 km

Dennis's average speed in the first two hours = 150 km/h

Dennis's average speed in the last two hours = 100 km/h

Therefore;

Dennis traveled 150 km/h × 2 h = 300 km in the first two hours

Dennis traveled 100 km/h × 2 h = 200 km in the last two hours

Which gives, Dennis traveled 300 km + 200 km in 2h + 2h = 4h

Therefore, Dennis traveled 500 km in 4 hours and Dennis must have stopped for an hour in the middle of this trip.

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2 years ago