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Masja [62]
3 years ago
5

Round off 53.96 to the nearest tenth​

Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

54.0

Step-by-step explanation:

53.96

The bold number is in the tenth form and needs rounding off. So look at the number after the bold number (6).

Now 6 falls off the 5 or greater category, meaning that you have to add one to the bold number.

9 + 1 = 10

Now since the number is 10, you add the number before the bold number by 1 (underlined number) and replace the bold number by 0.

5<u>3</u>.96

54.00

Since you are focusing on the tenth place, keep the zero sitting on that spot.

54.0

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Y=2x+4
the y intercept is 4 and the slope is 2/1
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The smiths bought an apartment for $75,000. assuming that the value of the apartment will appreciate at most 4% a year, how many
DedPeter [7]

Answer:

1st year is $75,000 * 0.04 = $3,000 $75,000+$3,000=$78,000

Step-by-step explanation:

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
In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses wer
Stella [2.4K]

Answer:

The number of houses built in Town B is 56.

Step-by-step explanation:

We are given that in 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B.

Also, 70 houses were built in Town A during 210.

Let the number of houses built in Town B be 'x'.

So, according to the question;

Number of houses built in Town A = Number of houses built in Town B + 25% of the houses built in Town B

70 = x + (25\% \times x)

70 = x(1+0.25)

x=\frac{70}{1.25}

x = 56

Hence, the number of houses built in Town B is 56.

8 0
3 years ago
Name a pair of complementary angles in the diagram above.
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Answer:

7&9

Step-by-step explanation:

Really, its just because, that's my explanation.

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