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natita [175]
3 years ago
11

Find the difference: 25 kg 32 g - 23 kg 83 g

Mathematics
1 answer:
TEA [102]3 years ago
5 0

Answer:

-kg^33 *(23g^51 - 25)

Step-by-step explanation:

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You can also write fractions as percents. To write a fraction as a percent, find an ________________________ ___________________
TiliK225 [7]

Answer:

common numerator and denomanator

Step-by-step explanation:

6 0
3 years ago
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HELP! the student council at woodlands middle school is planning a field trip. they polled randomly selected teachers and studen
gregori [183]

Answer:

c

Step-by-step explanation:

because on the chart is says 19% under burgers for teachers.

Tell me if it is right.

8 0
3 years ago
At the post office, 42% of packages ship priority mail. If 150 packages are shipped today, how many will not go priority mail?
alexdok [17]

(1 - 0.42)*150 = 63

0.58*150 = 87

87 packages were not priority mail.

3 0
3 years ago
Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin a
IgorLugansk [536]

Answer: -8.16 to 15.84

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval in which we are a percentage sure the true mean is in the interval.

A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be

x_{1}-x_{2} ± t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }

where

x₁ and x₂ are sample means

t is t-score

S_{p} is estimate of standard deviation

n₁ and n₂ are the sample numbers

The estimate of standard deviation is calculated as

S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

df=n_{1}+n_{2}-2

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }

S_{p}=\sqrt{\frac{299.07}{19} }

S_{p}= 15.74

The difference between means is

x_{1}-x_{2} = 17.5 - 13.66 = 3.84

Calculating the interval:

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{\frac{21}{108} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{0.194}

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.

5 0
3 years ago
Justin gave clerk $20 to pay a bill of $6.57. How much change should Justin get
hodyreva [135]
Well if he gave the clerk a $20 bill but paid $6.57 then the answer would be $13.43 because you will be subtracting $6.57 from $20
3 0
4 years ago
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