What is the ratio of salty snacks to sweet snacks in the vending machine

Mathematics

1 answer:

guajiro [1.7K]3 years ago

6 0

2859:682 is the ratio

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What is the equation of this line in standard form? −8x+7y=25 −9x+8y=−23 −8x+9y=−23 −8x+9y=23

kvasek [131]

<span>−8x+7y=25 is already in standard form, altho some people would prefer to re-write it as

</span><span>−8x+7y-25 = 0.

You have shared 4 equations here. Next time, please separate them with commas or semi colons, or type just 1 equation per line, for increased clarity. Thanks.</span>

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3 years ago

In Mr. Harold's class there are 12 students. They each have 5 pencils. Some of the students, s, each lost 2 pencils. Now, the st

Katena32 [7]

The answer would be C. (12 x 5) - (s x 2) = 38.

This is because if there are 12 students with 5 pencils per student, we can say 12 x 5. Then, think of the phrase, two pencils lost per student. That would be 2s, or (s x 2). Since it’s lost, we would subtract that amount from 12 x 5.

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3 years ago

Jocelyn boils 24 eggs. From experience, she knows that the shell of any given egg has a 0.04 probability of cracking during the

Varvara68 [4.7K]

Probability of an egg coming out of the boiling process with a cracked shell:
$1-(1-0.04)^{24} = 0.6246$

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2 years ago

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The cost for Andrea to produce several units of a certain item is $336 plus $12 per unit. She sells this item for $28 per unit.

Morgarella [4.7K]

Profit will be achieved when the money earned is greater than the money spent so,
336 + (n*12) = n*28 any subsequent units sold would result in profit
336 = 28n - 12n
336 = 16n
=> n = 336/16 = 21
21 units must be sold to draw even any more would result in profit

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3 years ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago