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jeka94
3 years ago
10

Write an inequality for each situation. Use x for the variable.

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0
1. x<8
2. x≥45
3. x≤38
4. x>85
You might be interested in
-9(5-2)-111÷3 show all work ​
Tasya [4]

Answer:

-74

Step-by-step explanation:

First: 5-2=3

-9*3=-27

111÷3=37

Then do -27-37=-74

3 0
3 years ago
Read 2 more answers
Maya's chocolate bar is 29% cocoa. If the weight of the chocolate bar is 67 grams, how many grams of cocoa does it contain? Roun
nika2105 [10]

Answer:

20

Step-by-step explanation:

x/67=29/100

29x67=1943

1943/100=

19.43

Round:

20

8 0
2 years ago
The graph of the function f(x)=3/x+5 is shown below.
loris [4]
The vertical asympote is the value of x that makes the denominator equal to zero:
x+5=0
Solving for x:
x+5-5=0-5
x=-5

The vertical asymptote is x=-5

Answer: Option A. x=-5
5 0
3 years ago
Read 2 more answers
In what year was her grandpa lopez twice as old as her dad?show how you know
RoseWind [281]
So,

We are trying to figure out when Grandpa Lopez's age was twice that of Dad.

Let x represent the number of years before/after when G. Lopez's age was twice that of Dad.

66 + x = 2(37 + x)

Distribute.
66 + x = 74 + 2x

Subtract x from both sides.
66 = 74 + x

Subtract 74 from both sides.
-8 = x

So 8 years ago, G. Lopez was twice as old as Dad.  Let's check that.

66 - 8 = 58

37 = 8 = 29

29 * 2 = 58
58 = 58

It checks.
8 0
3 years ago
You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a
Karo-lina-s [1.5K]

Answer:

573-1.690\frac{22}{\sqrt{36}}=566.8033    

573+1.690\frac{22}{\sqrt{36}}=579.1967      

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

\bar X=573 represent the sample mean

\mu population mean (variable of interest)

s=22 represent the sample standard deviation

n=36 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom are given by:

df=n-1=36-1=35

The Confidence level is 0.90 or 90%, the significance would be \alpha=0.1 and \alpha/2 =0.05, and the critical value for this case would be t_{\alpha/2}=1.690

And replacing we got:

573-1.690\frac{22}{\sqrt{36}}=566.8033    

573+1.690\frac{22}{\sqrt{36}}=579.1967    

And the 90% confidence interval would be between [566.8033 and 579.1967]

6 0
3 years ago
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