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Cerrena [4.2K]
3 years ago
7

If there are 6 grams of fat in a snack how many grams of protein are there?

Mathematics
1 answer:
vlabodo [156]3 years ago
7 0

6 grams of fat in a snack then 50 grams of protein in snacks

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the first four terms of a geometric progression are 2,10,50,250 a) write down the common ratio of the progression b) write down
VLD [36.1K]

Answer:

The common ratio is 5 and the 5th term is 1250

Step-by-step explanation:

The common ratio is given by

r=U(2)/U (1)

=10/2

=5

The fifth term

U (n)=ar^n-1

U(5)=2(5)^5-1

=2×5^4

=2×625

=1250

3 0
4 years ago
In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden
hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is \hat{p}=26/42=0.6190. The estimated standard deviation is given by \sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749. Because we have a large sample, the 90% confidence interval for p is given by 0.6190\pm z_{0.05}0.0749 where z_{0.05}=1.6448 is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 0.6190\pm (1.6448)(0.0749), i.e., (0.4958, 0.7422).

5 0
4 years ago
Jubal wrote the four equations below. He examined them, without solving them, to determine which equation has no solution.
umka2103 [35]

Answer:

3x + 2 = 3x - 2

Step-by-step explanation:

4 0
3 years ago
What is the largest square factor of 49896
alexira [117]

Answer:

The biggest factor of 49896 is 24948

3 0
3 years ago
Please helpp middle school math<br><br>suppose 4&lt;a&lt;7. find all the possible values of 15-2a​
bixtya [17]

9514 1404 393

Answer:

  1 < 15 -2a < 7

Step-by-step explanation:

There are a couple of ways you can do this.

1) Put the minimum and maximum values of a into the expression to see what its corresponding values are:

  15-2a for a=4:

     15-2(4) = 7

  15-2a for a=7:

     15-2(7) = 1

Then ...

  1 < 15-2a < 7

__

2) Solve for a in terms of the value of 15-2a, then impose the limits on a.

  x = 15 -2a

  2a = 15 -x

  a = (15 -x)/2

Now, impose the given limits:

  4 < (15 -x)/2 < 7

  8 < 15 -x < 14 . . . multiply by 2

  -7 < -x < -1 . . . . . . subtract 15

  7 > x > 1 . . . . . . . . multiply by -1

  1 < 15-2a < 7 . . . . . use x=15-2a

_____

The vertical extent of the attached graph is the range of possible values of 15-2a. It goes from 1 to 7.

8 0
3 years ago
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