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2 years ago

PLEASE HELP ASAP NEED NOW !!! For the given word problem, identify the rate of change.

Mathematics

1 answer:

lana [24]2 years ago

7 0

Answer:

6½ gallons

Step-by-step explanation:

Givem the cost of 1gallon of gas = $2.78. . If Cory buys $18 of gas, we are to find the number of gallons did he buy. To do that we will use the equality method as shown;

$2.78 ,= 1gallon

$18 = x gallon

Cross multiply

2.78 × x = 1×18

2.78x = 18

Divide through by 2.78

2.78x/2.78 = 18/2.78

.x = 6.5

Hence Cory buys about 6½ gallons of gas for $18

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HELP find angle relationships(picture included)

MariettaO [177]

Answer:

1. <I and <J are vertical angles

2. <G and <H form supplemantary angles

3. <G and <C are corresponding angles

4. <G and <F are alternate interiors

5. <I and <D are alternate exteriors

5 0

3 years ago

Use the raw data below to create a table that can be used to create a histogram with 5

schepotkina [342]

Answer:

Five classes

Step-by-step explanation:

1. Sort the data

You get

1.1, 1.3, 1.6, 1.7, 2, 2.6, 2.7, 2.9, 3.2, 3.2, 3.5, 3.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9, 4.9, 5.3, 5.7, 6.4, 6.5, 7.1, 7.5, 7.6, 8.1, 8.2, 9.2, 9.4

2. Calculate the range

Range = Max - Min = 9.4 - 1.1 = 8.3

3. Calculate the class width

Divide the range by the number of classes

8.3/5 = 1.7

Round this up to 2.

4. Decide where to start the histogram

You could use classes: 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 -11

However, even numbers are easier to read.

It would be preferable to use classes: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10

5. Prepare a frequency distribution table

Note: Each class does not include its largest possible value. Thus, a value of 2 goes into Class 2 - 4, (not Class 0 - 2).

$\begin{array}{cc}\textbf{Miles run} & \textbf{No. of students} \\0 - 2 & 4 \\2 - 4 & 9 \\4 - 6 & 8 \\6 - 8 & 5 \\8 - 10 & 4 \\\end{array}$

4 0

3 years ago

There are four activities on the critical path, and they have standard deviations of 1, 2, 4, and 2. what is the probability tha

Sauron [17]

The expected completion time is μ = 40 weeks.
The random variable, X = 38 weeks (the probable time)

If the standard deviation is σ, then the z-score is z = (x - μ)/σ.

Let us test the given standard deviations.
When σ=1,
z = (38-40)/1 = -2
From standard tables,
P(x<=38) = P(z<=-2) = 0.0228 =2.3%

When σ=2,
z = (38*40)/2 = -1
P(x<=38) = P(z<=-1) = 0.1587 = 15.9%

When σ=4,
z = (38-40)/4 = -0.5
P(x<=38) = P(z<=-0.5) = 0.3085 = 30.9%

Answers:
z=1 => 2.3%
z=2 => 16% (approx)
z=4 => 31% (approx)

7 0

2 years ago

Please somebody help me ASAP

Marat540 [252]

it's recorded that out of 1000 people, 762 wear the corrective lenses.

just divide 762 from 1000 and multiply that result by 100.

762/ 1000 = .762

.762 x 100 = ? %

which is 76.2 %

so, we predict that 76.2% of Americans would wear corrective lenses.

<h2>Answer: 76.2 %</h2>

7 0

2 years ago

Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card an

poizon [28]

Answer:

We have that $B = 0.3$ .

$0.3 = b + (A \cap B)$

However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.

Step-by-step explanation:

Event A:

Probability that a student has a Visa card.

Event B:

Probability that the student has a MasterCard.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student has a Visa card but not a MasterCard and $A \cap B$ is the probability that a student has both these cards.

By the same logic, we have that:

$B = b + (A \cap B)$

In this problem, we have that:

$A = 0.6, B = 0.3$

(a) Could it be the case that P(A ∩ B) = 0.5?

We have that $B = 0.3$ .

$0.3 = b + (A \cap B)$

However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.

8 0

3 years ago