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ohaa [14]
3 years ago
9

Enter the missing numbers in the boxes to complete the table of equivalent ratios

Mathematics
1 answer:
son4ous [18]3 years ago
4 0

Answer:

Time(min)  

2

6

8

15

Distance(km)

6

18

24

45

Step-by-step explanation:

the time is 1/3rd of the distance. it's a 1:3 ratio

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Three divided by four
muminat

Answer:

0.75

Step-by-step explanation:

3/4=0.75

Hope this helps! :)

8 0
3 years ago
John likes to hike during the summer he can usually hike about 3 miles every hour but when it's really hot he swims in a mountai
Nastasia [14]
During the 9 mile hike he  spends 2 periods of half hour swimming  = 1 hour.  
actual hiking time  = 3*3 = 9 hours

So total time = 9+1 = 10 hours
8 0
3 years ago
Help with volume all help is appreciated :)
Lina20 [59]

Answer:

It would be 0.41 ft^3

Step-by-step explanation:

Alright, to start, lets get the volume of the entire cinder block and the holes within it.

1.31 * 0.66 * 0.66 ---> 0.5706...

Next with the holes, they are both 0.33 wide, 0.39 long, and just as tall at 0.66 feet.

0.33 * 0.39 * 0.66 ---> 0.0849...

Since there's two of them: 0.1698...

To finish it, subtract the hole volume from the total volume;

0.5806 - 0.1698 = 0.4108

Rounds to <u>0.41 ft</u>^3

3 0
3 years ago
True or False: The range of a logarithmic function is always all real numbers.
Readme [11.4K]

Answer: true!

Step-by-step explanation: i do not know but true!

5 0
3 years ago
Read 2 more answers
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb
ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

6 0
3 years ago
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