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3 years ago

How do you graph n > -9?

Mathematics

1 answer:

Harman [31]3 years ago

3 0

Answer:

In order to graph this follow these steps

Put an OPEN dot on the number -9 and then draw or color a line to the right of the line.

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tekilochka [14]

Answer:

I think around 15°c

Step-by-step explanation:

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3 years ago

If the test for a disease is accurate 45% of the time, How often will it come back negative if a patient has the disease?

Dafna1 [17]

D. 55% of the time

45% of the time it will say positive

The other 55% of the time will come back negative.

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3 years ago

Six women push grocery carts up a ramp as shown.

Arturiano [62]

Six women push grocery carts up a ramp as shown.

2 carts weigh 50 lbs. each.

4 carts weigh 20 lbs. each.

How much work was done?

_____________ ft.-lbs. (total work)

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3 years ago

The difference of twice a number and 17 is 9

Papessa [141]

Answer:

Step-by-step explanation:

n = the number

2n - 17 = 9

2n = 26

n = 13

~Hope this Helps!~

3 0

3 years ago

Below are data on 18 people who fell ill from an incident of food poisoning. The data give the incubation period (the time in ho

Sveta_85 [38]

Answer:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

$Min =15$

$Q_1 = \frac{20+20}{2}=20$

$Median= Q_2= \frac{32+34}{2}=33$

$Q_3 = \frac{46+46}{2}=46$

$Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

Step-by-step explanation:

For this case we have the following data:

15 16 18 19 20 20 21 28 32 34 36 43 46 46 48 48 72 88

The sample size is n =18

We can calculate the mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

The minimum for this case is $Min =15$

Now we can find the 5 number summary.

For the first quartile we work with the first 10 observations: 15 16 18 19 20 20 21 28 32 34. And Q1 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_1 = \frac{20+20}{2}=20$

For the median since we have an even number for the sample size would be the average between the 9th and the 10th position from the dataset ordered and we got:

$Median= Q_2= \frac{32+34}{2}=33$

For the third quartile we work with the last 10 observations: 32 34 36 43 46 46 48 48 72 88. And Q3 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_3 = \frac{46+46}{2}=46$

The maximum is $Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

3 0

3 years ago