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miv72 [106K]
3 years ago
6

To practice for a competition, Carly swam 0.51 kilometer in the pool each day for 3 weeks. How many meters did Carly swim in tho

se 3 weeks?
1 km = 1,000 m
Mathematics
1 answer:
erica [24]3 years ago
7 0

Answer:

0.51km × 1000 = 510m

510m × (3×7) = 510m × 21 = 10710m in whole

Step-by-step explanation:

first we transform km to m (multiply by 1000)

then we count how much days are in three weeks ( 3 weeks × 7 days)

then we just multiply the meters per day × the days

99% it's true

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Honestly I’m so bad at geometry so helpppp:)
Anastasy [175]

Answer:

900 m

Step-by-step explanation:

l*w*h

12 m (l)*5 m (w)*15 m (h)= 900 m altogether

6 0
3 years ago
Making $40 payments each Friday,
Marianna [84]
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8 0
3 years ago
On each 3 days derrik rode 6.45 km to school, 150 meters to the library, and then 500 meters back home. How many kilometers did
Andrew [12]

Just add all 3 numbers up together:

6.45+ 0.15+0.5

=7.1

Since it was three days multiply the 7.1 by three

=7.1x3

=24 

So for three days derrick rode for 24 km 

Hoped I helped :)

7 0
3 years ago
A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di
cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10

Thus, a Normal approximation to binomial can be applied.

So,  p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n}).

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})

                   =P(Z>13.25)\\=1-P(Z

 *Use a <em>z</em> table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

3 0
3 years ago
You are building a rectangular planter for your school garden. You want the area of the bottom to be 90ft^2. You want the length
Anestetic [448]
15 feet and 6 feet
 Hope this helps you :)<span>
</span>
6 0
3 years ago
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