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salantis [7]
3 years ago
5

A student's pay of $18 an hour is $4 more than twice the amount the student earns per hour at an internship. Enter and solve an

equation to find the hourly pay of the internship.
Mathematics
1 answer:
Natasha2012 [34]3 years ago
4 0
1. You have that a<span> student's pay of $18 an hour </span>is <span>$4 more than twice the amount the student earns per hour at an internship.

 2. So, let's call the </span><span>hourly pay of the internship: "x". Then, you have:
</span><span>
 18=2x+4

 3. When you clear the variable "x", you obtain:

 18-4=2x
 14=2x
 x=14/2
 x=$7

 Therefore, the answer is: T</span><span>he hourly pay of the internship is $7.</span>
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Graph the system of equations on your graph paper to answer the question.
scZoUnD [109]
<span>y=x-4
y=-x+6 
---------------------
Substitute x - 4 for y in </span>y=-x+6 
x-4=-x+6<span>
--------------------------------
Add X on each side
</span><span>x-4+x</span>=-<span>x+6+x
</span>2x - 4 = 6
--------------------------------------
Add 4 on each side
<span>2x-4+4</span>=<span>6+4
</span>2x = 10
------------------------------------------------
Divide by 2 on each side
2x ÷ 2 = 10 ÷ 2
x = 5
Now we have X
================================================================
To find Y we substitute 5 for x in y=<span>x-<span>4
</span></span>y = 5 - 4
y = 1
-----------------------------------------------------
Your answers are Y = 1 and X = 5

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3 years ago
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8 0
4 years ago
27 players for 12 spots
labwork [276]

Answer: 27/12=2.25

Pretty simple:)

Step-by-step explanation:

7 0
3 years ago
Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

Where (from tables):

Z_{0.99}=2.33

Finally, the interval at 98% confidence level is:

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4 0
1 year ago
Which Factors would simplify in this expression?
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Answer:

(x+6)

Step-by-step explanation:

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