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3 years ago

7

Must a decimal divisor and a decimal dividend have the same number of decimal places in order to have a whole number quotient? W

rite a division equation using two decimal numbers to support your answer.

1 answer:

fredd [130]3 years ago

8 0

Your answer is B I’m pretty sure about that!

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F (x) = (x + 5)º(x - 9)(x + 1)<br> 3

JulijaS [17]

Answer:

3x^3 - 11x^2 + 93x - 105

Step-by-step explanation:

(x+5) (x-9) (x+1) (3) FOIL (First, Outside, Inside, Last)...

x^2 - 9x + 5x - 35 (3x+3) Multiply

3x^3 + 3x^2 - 27x^2 - 27x + 15x^2 + 15x - 105x - 105 Combine Like Terms...

3x^3 - 11x^2 + 93x - 105

8 0

2 years ago

Kelly inherits land which had a basis to the decedent of $95,000 and a fair market value of $50,000 on August 4, 2018, the date

Degger [83]

Answer:

The correct option is;

Her recognized <u>loss </u>is ($1,000)

Step-by-step explanation:

The given information are;

The basis of the land to the decedent = $95,000

The land's market value on 4th of August 2018 when the decedent died = $50,000

The alternate valuation date = 6 months + The date of death of the decedent = 4th February, 2019

The value filed by the executor on the tax return using the alternate valuation date = The market value of the estate on 4th of February 2019

The market value of the land on 4th of February 2019 = $45,000

∴ The value filed by the executor on the tax return using the alternate valuation date = $45,000

The value of the land on November 12, 2018 when the executor distributed the land to Kelly = $49,000

The value at which Kelly sells the land on June 10, 2019 = $48,000

Given that, recognized gain is the profit made from selling an asset based on the value of the asset when it was obtained, we have;

Kelly's recognized gain or loss = (The value at which Kelly sells the land) - (The value of the land when the executor distributed the land to Kelly)

Kelly's recognized gain or loss = $48,000 - $49,000 = -$1,000 = ($1,000)

Therefore, Kelly's recognized loss = ($1,000).

6 0

3 years ago

. The mean incubation time for a type of fertilized egg kept at a certain temperature is 25 days. Suppose that the incubation ti

Paul [167]

Answer:

Step-by-step explanation:

Given that the mean incubation time for a type of fertilized egg kept at a certain temperature is 25 days.

Let X be the incubation time for a type of fertilized egg kept at a certain temperature is 25 days.

X is N(25, 1)

a) Normal curve is in the attached file

b) the probability that a randomly selected fertilized egg hatches in less than 23 days

= $P(X$

we convert x into Z score and use std normal distn table to find probability

$P(X$

i.e. we can say only 2.5% proportion will hatch in less than 23 days.

8 0

3 years ago

I need help solving this!

blagie [28]

Answer:

miles hybrid car went = 9.80 gal × 54.1 miles/gal

= 530.18 miles

132km × 0.621 = 81.972 miles

530.18 miles = 9.80 gal

1 mile = 9.80/530.18

= 0.018 gal

81.972 miles = 0.018 × 81.972

= 1.515 gal

1.515 gal × 3.785 = 5.734 litres

3 0

3 years ago

One light-year equals 5.9 x 1012 miles. How many light-years are in 6.79

inna [77]

Option B

The number of light years in $6.79 \times 10^{16}$ miles is 11508 light years

<em><u>Solution:</u></em>

Given that,

One light-year equals 5.9 x 10^12 miles

Therefore,

$1 \text{ light year } = 5.9 \times 10^{12} \text{ miles }$

To find: Number of light years in $6.79 \times 10^{16}$ miles

Let "x" be the number of light years in $6.79 \times 10^{16}$ miles

Then number of light years in $6.79 \times 10^{16}$ miles can be found by dividing $6.79 \times 10^{16}$ miles by miles in 1 light year

$\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79 \times 10^{16}}{5.9 \times 10^{12}}\\\\\text{Use the law of exponent }\\\\\frac{a^m}{a^n} = a^{m-n}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79}{5.9} \times 10^{16-12}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 1.1508 \times 10^4\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 11508$

Thus number of light years in $6.79 \times 10^{16}$ miles is 11508 light years

3 0

3 years ago