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leva [86]
2 years ago
12

In January, the depth of a lakeksas 1,053 feet. In August, the depth of the lake was 737.1 feet. What is the percentage decrease

of
the depth of the lake from January to August?
OA. 43%
OB. 30%
OC. 36%
OD. 70%
Mathematics
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

<em>Your answer would be B. </em>30<em>%</em>

Step-by-step explanation:

To figure out the decrease, you would first subtract the two numbers.

1053 - 737.1 = 315.9

Next, you would multiply the number by 1053.

\frac{315.9}{1053} = .3

Then, you would multiply that by 100.

.3 * 100 = 30

<em>The answer would be </em>30<em>%.</em>

1053 * .3 = 315.9

1053-315.9 = 737.1

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Find the value of x in the isosceles triangle​
Ne4ueva [31]

Answer:

10

Step-by-step explanation:

Seeing the 5 12 13 triple you get 1/2x=5 so

x=10

4 0
3 years ago
Can I get some help with this question plzzz
victus00 [196]
All the possible combinations are
1-1 2-1 3-1 4-1 5-1 6-1
1-2. 2-2 3-2 4-2. 5-2. 6-2
1-3 2-3. 3-3. 4-3. 5-3. 6-3
1-4. 2-4. 3-4. 4-4. 5-4. 6-4
1-5. 2-5. 3-5. 4-5. 5-5. 6-5
1-6. 2-6. 3-6. 4-6. 5-6. 6-6

What is the total for each roll?

2. 3. 4. 5. 6. 7.
3. 4. 5. 6. 7. 8
4. 5. 6. 7. 8. 9
5. 6. 7. 8. 9. 10.
6. 7. 8. 9. 10. 11.
7. 8. 9. 10. 11. 12

What roll occurs most often?

>>>>>>7<<<<<

7 occurs in 6 out of 36 possible outcomes
7 0
3 years ago
Which of the following mixtures of paint could be used as the missing values in the table ? Choose 2 answers
Kazeer [188]

Answer:

weres the answer how am i going to answer this

Step-by-step explanation:

6 0
3 years ago
Scores on a test are normally distributed with a mean of 81.2 and a standard deviation of 3.6. What is the probability of a rand
Misha Larkins [42]

<u>Answer:</u>

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

<u>Solution:</u>

Given, Scores on a test are normally distributed with a mean of 81.2  

And a standard deviation of 3.6.  

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6  

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1

So, probability of getting more than 77.6 = 1 – area of z- score(-1)

= 1 – 0.1587 [Using z table values]

= 0.8413

Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185

Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.

4 0
3 years ago
The fit line equation is Y = -2.61x +152.51, where x is variable 1 and Y is variable 2. According to the equation, what is the e
egoroff_w [7]

The expected value of y when x is equals to 45 in the equation is 35.06

<h3>How to find variable from an equation?</h3>

The equation is given as follows;

y = -2.61x + 152.51

where

  • x = variable 1
  • y = variable 2

Therefore,

when

x = 45

y = -2.61x + 152.51

y = -2.61(45) + 152.51

y = - 117.45 + 152.51

y = 35.06

learn more on equation here: brainly.com/question/14279419

#SPJ1

5 0
2 years ago
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