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otez555 [7]
3 years ago
7

Shea makes $ 671.00 each week. how much does she make in one year?

Mathematics
2 answers:
azamat3 years ago
5 0

Answer:

34892

Step-by-step explanation:

671.00x52=$34892

k0ka [10]3 years ago
4 0
52 weeks in a year.

671 × 52 = 34892

Salary in 1 year: $34892.00
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Jared is using a 100 ft rope to set up a kite-shaped area for food vendors. He has started roping off the area as shown below, a
marishachu [46]
Jared has already used 30+30=60 ft of rope.

A kit shape has 2 pairs of 2 lines that are the exact same length. We can use x to represent this unknown length.

60+2x=100

Subtract 60 from both sides
2x=40

Divide both sides by 2
x=20

Final answer: He can use all the rope by using 20 ft for each of the other sides that are left.
8 0
3 years ago
Read 2 more answers
4. y = -(x+2)2 - 7<br><br>how do you write the equation in standard Form​
borishaifa [10]

Answer:

2x+4y=-11

Step-by-step explanation:

4. y = -(x+2)2 - 7

4y=-2x-4-7

4y=-2x-11

2x+4y=-11

3 0
3 years ago
Help how you solve this
Anettt [7]
Just add up how much Luna practiced which is 20+30+20 to get 70 and then add another 30 to get 100. So Elsa practiced for 100 minutes or 1hour 40 minutes.
8 0
3 years ago
0.25k+1.5-k-3.50.25k+1.5−k−3.5
BARSIC [14]

Answer:

-0.75k - 2.0

Step-by-step explanation:

0.25k + 1.5 − k − 3.5

Combining like terms with rational coefficients

0.25k + 1.5 - k - 3.5

Collect like terms with rational coefficients

=0.25k - k + 1.5 - 3.5

= -0.75k - 2.0

Answer equals negative zero Point seven five k minus two point zero

7 0
3 years ago
Assuming that the population is normally​ distributed, construct a 9090​% confidence interval for the population mean for each o
jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

 

Set B:

Mean: 4.5

Standard dev: 2.45

 

% =  90 

Set A:

Standard Error, SE = s/ √n =    1.9/√8 = 0.67  

Degrees of freedom = n - 1 =   8 -1 =  7   

t- score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

 

Set B:

Standard Error, SE = s/ √n =    2.45/√8 = 0.87  

Degrees of freedom = n - 1 =   8 -1 = 7   

t- Score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

 

<span>We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.</span>

6 0
3 years ago
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