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Ray Of Light [21]

3 years ago

8

At a recent party it cost $9.50 for refreshments for 10 guest.at this rate,how much would it cost to have refreshment for 80 gue

st?

1 answer:

Talja [164]3 years ago

8 0

For 80 guests you will need 80/10 = 8 times more than for 10 guests:

8*9.50 = $76

Refreshment for 80 guests costs $76.

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(×+2) +(4×-122)=180 need help with this problem

alina1380 [7]

So first you need to open the brackets, so it would be x+2+4x-122=180. Then we can add the 4x and the other x to make 5x, and then by doing 2-122 we get -120. This gives us the equation 5x-120=180. We then isolate the variable by moving the -120 to the other side of the equation and becoming a positive, so it would look like 5x=180+120. Then, we have 5x=300. 300 divides by 5 is 60 making X=60. Hope this helps!

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2 years ago

Why do you think the Court ruled differently in New York Times Co. v. United States (1971) than it did in Schenck v. United Stat

Marina CMI [18]

Answer:

The case reached the Supreme Court in June 1971. The Supreme Court has, at times, ruled that the government can restrict speech that presents a “clear and present danger.”

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3 years ago

Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor

iren2701 [21]

Answer:

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,0.08)$

Where $\mu$ and $\sigma=0.08$

Since the distribution for X is normal then the we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

$SE= \frac{0.08}{\sqrt{24}}= 0.0163$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,0.08)$

Where $\mu$ and $\sigma=0.08$

Since the distribution for X is normal then the we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

$SE= \frac{0.08}{\sqrt{24}}= 0.0163$

8 0

3 years ago

Ospreys are birds that catch and eat fish. An osprey dived from a height of 36 feet above the surface of a lake to a

nikitadnepr [17]

Answer:

C

Step-by-step explanation: 36 is the height. Minus 4 is 32. So ye...

7 0

2 years ago

Two cargo ships spot a signal fire on a small island. The captains know they are 140 feet

Irina18 [472]

The distance from Ship B to the signal fire at point C is $405.4 \mathrm{ft}$

Explanation:

It is given that two ships A and B are at a distance 140 ft from each other.

The angle formed by ship A is 82º and the angle formed by ship B is 78º.

To determine the distance from ship B to the signal fire at point C, we need to know the another angle.

Hence, we have,

$\begin{aligned}\angle A+\angle B+\angle C &=180^{\circ} \\82^{\circ}+78^{\circ}+\angle C &=180^{\circ} \\160^{\circ}+\angle C &=180^{\circ} \\\angle C &=20^{\circ}\end{aligned}$

Now, we shall solve the problem using the law of sines,

$\frac{\sin A}{a}=\frac{\sin C}{c}$

Substituting the values, we have,

$\frac{\sin 82^{\circ}}{a}=\frac{\sin 20^{\circ}}{140}$

$\frac{0.9903}{a}=\frac{0.3420}{140}$

Simplifying, we get,

$a=\frac{0.9903 \times 140}{0.3420}$

$a=405.4$

Thus, the distance from Ship B to the signal fire at point C is $405.4 \mathrm{ft}$

4 0

2 years ago