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sweet [91]
4 years ago
12

Please answer this correctly

Mathematics
2 answers:
natali 33 [55]4 years ago
8 0

Answer: 30

Step-by-step explanation:

Q1: 120

Q3: 150

To find the interquartile range, subtract Q1 from Q3, which is 150-120. Therefore, the interquartile range  of the kitten's weight, is 30

Mnenie [13.5K]4 years ago
5 0

Answer: 30 grams

Step-by-step explanation:

The interquartile range is the range within the boxed areaa. You subtract the minimum value from the maximum value.

150 - 120 = 30

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Marat540 [252]

Answer:

y= 5x -2 or y-13 = 5 (x-3)

Step-by-step explanation:

The first one is slope-intercept form, the second is point-slope form

3 0
4 years ago
Three different accounts are described below. Order the accounts according to their values after 10 years, from greatest to leas
faust18 [17]

Answer:

`1. You deposit $950 in an account that earns 9% annual interest compounded daily.

2.You deposit $1000 in an account that earns 8% annual interest compounded daily.

3. You deposit $1000 in an account that earns 9% annual interest compounded semiannually.

Step-by-step explanation:

Trick question, number 3 says semiannually so it is only twice a year, so it is the least. In math papa or any algebra calculator, put in the two equations and compare which is greater, in this case, y = 950(1.09)^10 is greater than y=1000(1.08)^10

3 0
3 years ago
Emma is doing a science project. She is measuring the time it takes a snail to move 5 kilometres. It takes the snail one hour to
san4es73 [151]

Answer:

41.67hrs

Step-by-step explanation:

if the snail moves 0.12km in 1hr

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4 0
3 years ago
Read 2 more answers
colby and jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the num
kaheart [24]
To get started, we will use the general formula for bacteria growth/decay problems:

A_{f} =  A_{i} ( e^{kt} )

where: 
A_{f} = Final amount
A_{i} = Initial amount
k = growth rate constant
t = time


For doubling problems, the general formula can be shortened to:

kt = ln(2)

Now, we can use the shortened formula to calculate the growth rate constant of both bacteria:

Colby (1):
k_{1} = ln(2)/t
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Jaquan (2):
k_{2} = ln(2)/t
k_{2} = ln(2)/3 = 0.23105 per hour

Using Colby's rate constant, we can use the general formula to calculate for Colby's final amount after 1 day (24 hours).

Note: All units must be constant, so convert day to hours.

A_{f1} = 50( e^{0.34657(24)})
A_{f1} = 204,800

Remember that the final amount for both bacteria must be the same after 24 hours. Again, using the general formula, we can calculate the initial amount of bacteria that Jaquan needs:

A_{f2} = 204,800 =  A_{i2} ( e^{0.23105(24)} )
A_{i2} = 800

3 0
3 years ago
1/2 of 3 inches is what inches<br><br><br> ​
muminat

Answer:

1.5 inches

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half of 3 is 1  1/2 or 1.5

7 0
3 years ago
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