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tamaranim1 [39]
3 years ago
7

A rectangular solid has sides of 7 cm, 9 cm, and 11 cm. What is its surface area?

Mathematics
1 answer:
otez555 [7]3 years ago
7 0

Answer:

Surface area is, 478 square centimeter

Step-by-step explanation:

Surface area (S)of rectangle is given by:

S = 2(lw+wh+hl)                 .....[1]

where,

l is the length

w is the width

h is the height of the rectangle respectively.

As per the statement:

A rectangular solid has sides of 7 cm, 9 cm, and 11 cm

⇒ l = 7 cm , w = 9 cm and h = 11 cm

Substitute in [1] we have;

S = 2(7 \cdot 9+9 \cdot 11+11 \cdot 7)

⇒S = 2(63+99+77) = 2 \cdot 239 = 478 cm^2

Therefore, the surface area is, 478 square centimeter

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Step-by-step explanation:

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3 years ago
Find ln(-127.1).
cluponka [151]

Answer:

The natural logarthim of a negative is undefined.

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3 years ago
Using the midpoint method, calculate the price elasticity of demand of Good Z using the following information: When the price of
exis [7]

Answer: (B) The price elasticity of demand for good Z = 0.86

Step-by-step explanation:

The formula for determining elasticity of demand by using the midpoint method is

(Q2 - Q1)/[(Q2 + Q1)/2] / (P2 - P1)/[(P2 + P1)/2]

Where

P1 is the initial price of the item.

P2 is the final price of the item.

Q1 is the initial quantity demanded for the item.

Q2 is the final quantity demanded for the item.

From the information given,

P1 = 10

P2 = 15

Q1 = 85

Q2 = 60

The price elasticity of demand for good Z = (60 - 85)/[(60 + 85)/2] / (15 - 10)/[(15 + 10)/2]

= (-25/72.5) / (5/12.5) = -25/72.5 × 12.5/5

= - 312.5/362.5 = - 0.86

6 0
4 years ago
Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif
dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals t_{0.01} =2.82  (from tabulated value).

The mean value = 0.0365

x±t_{0.01} \frac{S}{\sqrt{n-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{22} }

0.0365±2.82 \frac{0.0046}{4.690 }

using calculator

0.0365±0.00276

Confidence interval is

(0.0365-0.00276,0.0365+0.00276)

(0.0338,0.0392)

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

4 0
3 years ago
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