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Serga [27]
2 years ago
14

) How long does he travel at 30 km/h, if he travels 1.2 h at 20 km/h?​

Mathematics
1 answer:
GREYUIT [131]2 years ago
4 0

Step-by-step explanation:

please mark me as brainlest

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Factor 10c^2-60cd +80d^2
ozzi
10c^2-60cd +80d^2

(2c - 8d) (5c -10d)

this  \\ is \\ because...

2c*5c=10c^2

2c * -10d=-20cd \\ -8d*5c=-40cd \\ -20cd-40cd=-60cd

-8d * -10d = 80d
5 0
3 years ago
1. 5% of $30 =
stira [4]

Answer:

1 is 1.5

2. is 15

3 is 0.4

4. is 3

5. is 0.8

6 is 10

Step-by-step explanation:

Good luck!

3 0
3 years ago
If you can help me with Slope Intercept plz help me
11Alexandr11 [23.1K]

Answer:

y = 6/5x - 2 (or y = 1.2x - 2)

Step-by-step explanation:

To find the slope-intercept form ( = mx + b) you need to rearrange the equation of the line:

Step 1 - Add 18x to both sides

15y = 18x - 30

Step 2 - Divide both sides by 15.

y = 6/5x - 2

4 0
3 years ago
3-(12-6÷3)+7(3×2-5)​
insens350 [35]

Answer:

Solutions

Step-by-step explanation:

3-(12-2)+7(3*2-5)

3-10+7(6-5)

3-10+7*1

3-17

-14

8 0
3 years ago
Read 2 more answers
There are 255 students this term completing this same assignment. Assuming they calculated the CI correctly, how many students s
andrey2020 [161]

Answer:

The number of students we expect to have an interval that does not contain the true mean value is,  255\times [\alpha\%].

Step-by-step explanation:

A [100(1 - α)%] confidence interval for true parameter implies that if 100 confidence intervals are created then [100(1 - α)] of these 100 confidence intervals will consist the true population parameter value.

Here α is the significance level. It is defined as the probability rejecting the claim that the true parameter value is not included in the 100(1 - α)% confidence interval.

It is provided that 255 students create the same confidence interval, correctly.

Then the number of students we expect to have an interval that does not contain the true mean value is,  255\times [\alpha\%]

For instance, if the students are creating a 95% confidence interval for mean then the number of students we expect to have an interval that does not contain the true mean will be:

The significance level is:

Confidence\ level=100(1-\alpha)\\\frac{95}{100}=(1-\alpha)\\\alpha =1-0.95=0.05

Number of students we expect to have an interval that does not contain the true mean will be: 255\times [\alpha\%]=255\times 0.05=12.75\approx13

Thus, 13 of the 255 confidence intervals will not consist the true mean value.

3 0
3 years ago
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