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Lisa [10]
3 years ago
13

Jorge is scheduled to work 36 hours this week. He has already worked 27.25 hours. How much more hours does gorge have to work th

is week?
Mathematics
1 answer:
lesantik [10]3 years ago
6 0

Answer:

Step-by-step explanation:

Beth earns $54 per day and $10 for each extra hour she works. Ray earns $60 per day and $8 for each extra hour he puts in. They both work five days a week. The equations show their weekly earnings with respect to how many extra hours they work.

Beth: y = 270 + 10x

Ray: y = 300 + 8x

This system is graphed below.

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Classify the triangle. Select all that apply
dlinn [17]
One of the answers is equilateral because all of the sides are of equal length. 
The other answer is acute because all of the angles are under 90º since they are all 60º. Hope this helps!
6 0
3 years ago
Read 2 more answers
Analyze the key features of the graph f(x) = 1/x+3 -2
tia_tia [17]
<h2>Explanation:</h2><h2></h2>

Here we have the following rational function:

f(x) = \frac{1}{x+3} -2

So the graph of this function is shown in the First Figure below. Let's define another function which is a parent function:

g(x)=\frac{1}{x}

Whose graph is shown in the second figure below. So we can get the graph of f from the graph of g this way:

Step 1. Shift the graph 3 units to the left:

f_{1}(x) = \frac{1}{x+3}

Step 2. Shift the graph 2 units down:

f(x) = \frac{1}{x+3}-2

Finally, the features of the graph of f are:

The graph of this function comes from the parent function g and the transformations are:

  • A shifting 3 units to the left
  • A shifting 2 units down

8 0
3 years ago
Listed below are the overhead widths​ (in cm) of seals measured from photographs and the weights​ (in kg) of the seals. Construc
Anarel [89]

Answer and Step-by-step explanation: Scaterplot is a type of graphic which shows the relationship between to variables. In this question, you want to determine if there is a linear relationship between overhead widths of seals and the weights. So, the hypothesis are:

H₀: no linear correlation;

H₁: there is linear correlation;

In this hypothesis test, to reject H₀, the correlation coefficient r of the data set has to be bigger than the critical value from the table.

With α = 0.05 and n = 6, the critical value is 0.811.

The linear correlation is calculated as:

r = n∑xy - ∑x.∑y / √[n∑x² - (∑x)²] [n∑y² - (∑y)²]

r = \frac{6.9632 - 51.1108}{\sqrt{6.439 - (51^{2} )}.6.14482 - (1108^{2} ) }

r = 0.9485

Since r is bigger than the critical value, H₀ is rejected, which means there is enough evidence to conclude that there is linear correlation between overhead widths and the weights.

In the attachments is the scaterplot of the measurements, also showing the relationship.

5 0
4 years ago
Given the function F (x) = x2 -2x -5 determine the average rate of change of the function over the interval -5 &lt; x &lt;6.
KengaRu [80]

Answer:

1

Step-by-step explanation:

Here f(x) = x^2 - 2x - 5, which at x - -5 is 25 +10 - 5 = 30 and at x = 6 is 19.

The average value of a function f(x) over an interval [a, b] is

                  f(b) - f(a)

ave. val. = ---------------

                     b - a

which in this particular case is

                     19 - 30

ave. val. = ----------------- = -11/11 = 1

                     6 - (-5)

The average value of this function f(x) = x^2 - 2x - 5 on [-5, 6] is 1.

4 0
3 years ago
Let X denote the time in minutes (rounded to the nearest half minute) for a blood sample to be taken. The probability mass funct
Basile [38]

This question is incomplete, the complete question is;

Let X denote the time in minutes (rounded to the nearest half minute) for a blood sample to be taken. The probability mass function for X is:

x     0       0.5       1       1.5       2       2.5

f(x)  0.1     0.2     0.3    0.2     0.1       0.1

determine;

a) P( X < 2.5 )

B) P( 0.75 < X ≤ 1.5 )

Answer:

a) P( X < 2.5 ) = 0.9

b) P( 0.75 < X ≤ 1.5 ) = 0.5

Step-by-step explanation:

Given the data in the question;

The probability mass function for X is:

x     0       0.5       1       1.5       2       2.5

f(x)  0.1     0.2     0.3    0.2     0.1       0.1

a) P( X < 2.5 )

P( X < 2.5 ) = p[ x = 0 ] + p[ x = 0.5 ] + p[ x = 1 ] + p[ x = 1.5 ] + p[ x = 2 ]

so

P( X < 2.5 ) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1

P( X < 2.5 ) = 0.9

b) P( 0.75 < X ≤ 1.5 )

P( 0.75 < X ≤ 1.5 ) = p[ x = 1 ] + p[ x = 1.5 ]

so

P( 0.75 < X ≤ 1.5 ) = 0.3 + 0.2

P( 0.75 < X ≤ 1.5 ) = 0.5

3 0
3 years ago
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