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zheka24 [161]
2 years ago
10

The temperature in fargo was -28 degrees on monday afternoon. by the evening it dropped an additional 15 degrees. what was the t

emperature in the evening
Please this is due in 5 minutes do it quick with step by step
Mathematics
1 answer:
sashaice [31]2 years ago
7 0

Step by step explanation: -28-15

When adding negative with positive it will always be negative so for this question, subtract -28 with 15 which would be -43. So -43 would be the temperature in the evening.

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The relationship between the number of pencils and the number of pens in Sophie’s pencil case can be represented using the ratio
tekilochka [14]

ratio of 6:10, if you divide both numbers by 2 you get a ratio of 3:5

multiply by 2 you get 12:20

multiply by 1.5 you get 9:15


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3 years ago
Find the arrival time if the plane took off at 8:35 A.M. and flew for 2 hr. 55 min. Include A.M. or P.M. in your answer.
FrozenT [24]
11:30 am. 8:35 + 2 hours = 10:35 + 5 mins = 10:40 + 50 mins = 11:30 am
3 0
3 years ago
write the equation of the line that is perpendicular to the line y=-1/5x+9 and passes through the point (-2,-2)
OLEGan [10]

Answer:y=5x+8

Step-by-step explanation: the perpendicular slope is going to be the negative reciprocal of your given slope. In this case the perpendicular line has a slope of 5, because u flip -1/5 to make -5 and u negate -5 to make 5. This perpendicular slope passes through (-2,-2) and u find the y intercept like this: -2=(5(-2))+b. Therefore the y intercept is 8 and the perpendicular slope is y=5x+8

4 0
3 years ago
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katovenus [111]

Answer:

The car gets 30 miles to the gallon.  After the car has traveled 80 miles , 2 2/3 gallons have been consumed.

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5 0
3 years ago
(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each
Leviafan [203]

Following are the solution parts for the given question:

For question A:

\to (n) = 16

\to (\bar{X}) = 410

\to (\sigma) = 40

In the given question, we calculate 90\% of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}

\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\

Using the t table we calculate t_{ \frac{\alpha}{2}} = 1.753  When 90\% of the confidence interval:

\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53

So 90\% confidence interval for the mean weight of shipped homemade candies is between 392.47\ \ and\ \ 427.53.

For question B:

\to (n) = 500

\to (X) = 155

\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31

Here we need to calculate 90\% confidence interval for the true proportion of all college students who own a car which can be calculated as

\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}

\to C.I= 0.90

\to (\alpha) = 0.10

\to \frac{\alpha}{2} = 0.05

Using the Z-table we found Z_{\frac{\alpha}{2}} = 1.645

therefore 90\% the confidence interval for the genuine proportion of college students who possess a car is

\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344

So 90\% the confidence interval for the genuine proportion of college students who possess a car is between 0.28 \ and\ 0.34.

For question C:

  • In question A, We are  90\% certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
  • In question B, We are  90\% positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:  

brainly.in/question/16329412

7 0
2 years ago
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