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Inga [223]
3 years ago
12

lele's soccer team scored a total of 40 goals this year. They scored 15 more goals this year than they did last year. Write an a

lgebraic equation that could be used to represent g, the number of goals team scored last year.
Mathematics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:

g + 15 = 40

or

40 - 15 = g

Step-by-step explanation:

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A rectangular garden has length three times as great as its width. A second rectangular garden has the same width as the first g
evablogger [386]
Let x represent the width of the garden(s). Then the length of the second garden is (3x+3). The area is the product of length and width,
  90 = x(3x+3)
  30 = x(x+1)
At this point, you can recognize that 5 and 6 are factors of 30 that will satisfy this equation, or you can solve the quadratic
  x² +x -30 = 0
  (x +6)(x -5) = 0
  x = 5 . . . . . . . the negative solution does not apply

The width of the garden is 5 m.
6 0
3 years ago
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
4 years ago
What is the equivalent fraction for 4\7
Bingel [31]
4/7 = 8/14 = 12/21 = 16/28, etc
7 0
4 years ago
Read 2 more answers
Find the value of the expression. <br> y3 + x <br> for x = 6 and y = 1
andrezito [222]

Answer:

7

Step-by-step explanation:

Substitute the value of the variable into the expression and simplify

4 0
4 years ago
Read 2 more answers
A salesman receives a weekly salary of $150. He
shusha [124]

Answer:

150 + 4x = 330

4x = 180

x = 45

Step-by-step explanation:

8 0
3 years ago
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