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3 years ago

If a company uses $1,410 of its cash to purchase supplies, the effect on the accounting equation would be:

1 answer:

4 0

The above mentioned transaction will affect the Assets side of the accounting equation or the Assets side of the balance sheet

Since we use cash of $1,410 the cash account would be debited with $1,410. In other words cash will be reduced by $1,410

Since the cash is used to purchase supplies, the inventory will be credited with $1,410. This means the inventory would be increased by $1,410.

Hence in conclusion, the two accounting activities would be a debit of $1,410 on cash and a credit of $1,410 on the inventory. So the net of the Assets remains unchanged.

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Carroll borrowed $500 at 14% interest for 6 months. How much interest did she pay?

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3 years ago

Solve.<br><br>4×[52×(3×15)+32]-5=

worty [1.4K]

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3 years ago

The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (μ) earned by health-d

Jobisdone [24]

Missing Part of Question

Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $8,700

........

Answer:

z = 1.157

Step-by-step explanation:

Given

Annual profit as calculated = $90,300.

H0: μ = 89,000

Ha: μ > 89,000

σ = 8700

n = 60

Calculate the test statistic (z) that corresponds to the sample and hypotheses.

Test statistic (z) is calculated by

z = (x - μ)/(σ/√n)

Where x = 90,300

μ = 89,000

σ = 8700

n = 60

z = (90,300 - 89,000)/(8700/√60)

z = 1.157443298866584

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3 years ago

A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position

pantera1 [17]

Answer:

233.48s

3.84 min

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.

We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:

$y=A cos(\omega t+\phi)+C$

where:

A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

$\omega=$ angular speed.

in this case:

$\omega=\frac{2\pi}{T}$

where T is the period, in this case 6 min or

$6min(\frac{60s}{1min})=360s$

so:

$\omega=\frac{2\pi}{360}$

$\omega = \frac{\pi}{160}$

and

$\phi$ = phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

$y=-20 cos(\frac{\pi}{180}t)+22$

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

$-20 cos(\frac{\pi}{180}t)+22>13$

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

$-20 cos(\frac{\pi}{180}t)+22=13$

and solve it:

$-20 cos(\frac{\pi}{180}t)=13-22$

$-20 cos(\frac{\pi}{180}t)=-9$

$cos(\frac{\pi}{180}t)=\frac{9}{20}$

and we can take the inverse of cos to get:

$\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})$

which yields two possible answers: (see attached picture)

$\frac{\pi}{180}t=1.104$ or $\frac{\pi}{180}t=5.179$

so we can solve the two equations. Let's start with the first one:

$\frac{\pi}{180}t=1.104$

$t =1.104(\frac{180}{\pi})$

t=63.25s

and the second one:

$\frac{\pi}{180}t=5.179$

$t=5.179(\frac{180}{\pi})$

t=296.73s

so now we can build our possible intervals we can use to test the inequality:

[0, 63.25] for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25] for a test value of 1

$-20 cos(\frac{\pi}{180}(1))+22>13$

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

$-20 cos(\frac{\pi}{180}(70))+22>13$

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

$-20 cos(\frac{\pi}{180}(300))+22>13$

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

$233.48(\frac{1min}{60s})=3.84 min$

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

3 0

2 years ago