The amount invested at 11% is $5,000
The amount invested in stock is $14,000
What is the net profit on both investments?
The profit of each investment is the rate of return or loss multiplied by the amount invested
Let us assume that x was invested at 11% and the remaining 19000-x was invested at a loss rate of 3%
net profit=(11%*x)+(19000-x)*-3%
net profit=130
130=(11%*x)+(19000-x)*-3%
130=0.11x-570+0.03x
130=0.14x-570
130+570=0.14x
700=0.14x
x=700/0.14
x=$5,000
Amount invested in stock=19000-x
Amount invested in stocks=19000-5000
Amount invested in stocks=$14,000
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Answer:
Reduce the expression, if possible, by cancelling the common factors.
Exact Form:
11/20
Decimal Form:
0.55
22/40
Factor 2 out of 22.
2(11)/40
Cancel the common factors.
Tap for more steps...
11/20
The result can be shown in multiple forms.
Exact Form:
11/20
Decimal Form:0.55
Step-by-step explanation:
Hope it is helpful....
Answer:
b = −108
Step-by-step explanation:
Solve for b by simplifying both sides of the equation, then isolating the variable.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
Education : 11 11 8 13 17 11 11 11 19 13 15 9 15 15 11
Internet use 10 5 0 14 24 0 15 12 20 10 5 8 12 15 0
Labeled scatter plot of Education and Internet Use is attached in the picture below.
Yes there appears to be a linear relationship between the two variables (Education and Internet Use) as the data points appears to have an upward trend depicted by the linear trend line in the graph.
The correlation Coefficient value which is measures the degree of linear relationship between education and internet use is 0.7048
Answer: " 94. 6 ft² ".
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<u>Step-by-step explanation</u>:
The formula for the area of a trapezoid:
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A = (1/2) * (b1 + b2) * (perpendicular height, "h" ) ;
in which:
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"A = area (in units of ft²) "
"b1 = length of (base one) = 7.7 ft (from diagram) ;
"b2 = length of (base two—the other base) = 14.3 ft. ;
"h = height (perpendicular) = 8.6 ft.
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To solve for the Area, "A" ; we plug in the known values:
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A = (1/2) * (b1 + b2) * (perpendicular height, "h" );
= (1/2) * (7.7 ft + 14.3 ft) * (8.6 ft) ;
= (1/2) * (22 ft.) * (8.6 ft) ;
= (11 ft.) * (8.6 ft.) ;
= (11) * (8.6) * (ft.) * (ft.) ;
= (11) * (8.6) * ft² ;
A = " 94. 6 ft² ".
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Hope this is helpful to you!
Best wishes!
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