Mark Pryor, who lives in the Eastern Time Zone, placed a phone call at 7:00 A.M. to his uncle, who lives in the Pacific Time Zone. What time was it in the Pacific Zone when he called? Include A.M. or P.M. in your answer.

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Answer 1
Answer: In the Pacific Zone was 4: 00 A.M. when he placed the phone call.

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What are the three elements to calculate simple interest

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Simple interest is found by multiplying the amount of money deposited, the rate ( like .05 per month ) , and the time it will be in the bank ( one moth, 5 months, a year, etc. ).

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Mr. Li records the measures of the lengths of his students’ handprints. The lengths, in centimeters, are shown in the table.14.0 , 11.5 , 12.1 , 16.2 , 13.5 , 14.3 , 16.8
12.4 , 13.7 , 12.0 , 14.7 , 15.2 , 11.9 , 15.6
13.8 , 14.2 , 12.5 , 15.0 , 16.0 , 13.1 , 11.7

If the class creates a histogram of the data in the table, how many students are in the range 12 cm to 13.9 cm?

1) 3
2) 4
3) 7
4) 8

Answers

The correct answer is:

4) 8

Explanation:

We want to find the frequency of this range. To do this, we count how many values fall between 12.0 and 13.9. There are 8 total values that do this, so the answer is 8.

In the given data set, there are 7 values in the given range.

How many students are in the range 12 cm to 13.9 cm?

Here we just need to find the number of values between 12cm and 13.9 cm, from the given table, these are:

12.1 cm, 13.5 cm, 12.4 cm, 13.7 cm, 13.8 cm, 12.5 cm, 13.1 cm

So there are 7 values in the given range, which means that 7 students are in the range of 12cm and 13.9cm.

Then the correct option is the third one.

If you want to learn more about data sets:

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William wants to hang a mirror in his room. The mirror and frame must have an area of 50 square feet. The mirror is 5 feet wide and 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?3x2 + 13x − 20 = 0
4x2 + 14x − 7 = 0
4x2 + 26x − 10 = 0
2x2 + 2x − 10 = 0

Answers

The last one. 2x2+2x-10=0

If ∠BAC = 17° and ∠CED = 17° are the two triangles, ΔBAC and ΔCED similar? If so, by what criterion?A) yes, by AA similarity criterion
B) yes, by SAS similarity criterion
C) yes, by SSA similarity criterion
D) no, not possible to tell.

Answers

Answer:

Option A is correct.

Yes, by AA similarity criterion

Step-by-step explanation:

AA(Angle-Angle) Similarity Criterion states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

In ΔBAC and ΔCED

\angle BAC = \angle CED = 17^(\circ)   [Angle]        [Given]

\angle C = \angle C     [Angle]          [Common angle]

Therefore, by AA similarity criterion;

\triangle BAC \sim \triangle CED

Therefore,  ΔBAC and ΔCED are similar by AA similarity criterion.

If ∠BAC = 17° and ∠CED = 17° are the two triangles, ΔBAC and ΔCED similar? If so, by yes, by AA similarity criterion 

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Need help in this question please show work along with answer

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                              (x² - y²)² + (2xy)² = (x² + y²)²
                     (x² - y²)(x² - y²) + 2x²y² = (x² + y²)(x² + y²)
            x²(x² - y²) - y²(x² - y²) + 2x²y² = x²(x² + y²) + y²(x² + y²)
x²(x²) - x²(y²) - y²(x²) + y²(y²) + 2x²y² = x²(x²) + x²(y²) + y²(x²) + y²(y²)
                x⁴ - x²y² - x²y² + y⁴ + 2x²y² = x⁴ + x²y² + x²y² + y⁴
                       x⁴ - 2x²y² + y⁴ + 2x²y² = x⁴ + 2x²y² + y⁴
                       x⁴ - 2x²y² + 2x²y² + y⁴ = x⁴ + 2x²y² + y⁴
                                              x⁴ + y⁴ = x⁴ + 2x²y² + y⁴
                                            - x⁴         - x⁴
                                                     y⁴ = 2x²y² + y⁴
                                                   - y⁴              - y⁴
                                                      0 = 2x²y²
                                                      2       2
                                                      0 = x²y²
                                                      0 = xy

I didn't find the answer to the problem.

If 1 gallon equals 3.8 liters, how many liters are in 12 gallons? Round your answer to the nearest tenth.

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