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3 years ago

How do you solve 821*54 using an area model

Mathematics

1 answer:

SpyIntel [72]3 years ago

8 0

Answer:

800 20 1

50| | | |

4 | | | |

pretty much a rectangle with 800 20 1 at the top

50 4 on the side

and split the numbers in the box - i tried to recreate it as much as possible

oh then in each little box you have to multiply the numbers so

50 x 800 = 40,000 in the box

50 x 20 = 1,000 in the next box

50 x 1 = 50 in the next box

4 x 800 = 3,200

4 x 20 = 80

4 x 1 = 4

then you add all of them which would be 44,334

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Evaluate if a= -21 a/3+a=

ioda

The answer is -28
(sorry if i get this wrong)

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3 years ago

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7xy + 8z - 2z + 12xy + 10z

cestrela7 [59]

Answer:

19xy + 16z

Step-by-step explanation:

combine like terms

7xy +1 2xy= 19xy

8z - 2z + 10z = 16z

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3 years ago

What is the slope of the line shown in the graph? (4 points)

Korvikt [17]

Answer:

a. or $\frac{3}{-2}$

Step-by-step explanation:

Slope is equal to y1-y2 over x1-x2.

So the equation would be:

$\frac{2-(-1)}{-3-(-1)}$

2-(-1)=2+1=3

-3-(-1)=-3+1=-2

So the answer is:

$\frac{3}{-2}$

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3 years ago

What is the sum of 6dm, 40mm, 2m and 10cm in meters?

sveticcg [70]

Answer:

Step-by-step explanation:

convert to meters

6 dm = 0.6 m

40 mm = 0.04 m

2 m = 2 m

10 cm = 0.1 m

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0.6 + 0.04 + 2 + 0.1 =

2.74 m

7 0

2 years ago

Read 2 more answers

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago