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2 years ago

Anyone know how to do this?!!

Mathematics

1 answer:

iren2701 [21]2 years ago

6 0

Answer:

slope is -1/4

Step-by-step explanation:

$\frac{-4}{12}$ = $\frac{-1}{4}$

$\frac{-2}{8}$ = $\frac{-1}{4}$

and so on...

formula: $\frac{x}{y}$

hope this helps!!!

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On average, Mackenzie drinks of an 8-ounce glass of water in 24 hours. How much

vichka [17]

Answer:

1/3

Step-by-step explanation:

If Mackenzie drinks 8 ounces of water per 24 hours, that could be written as a ratio and a fraction. 8:24, 8/24

8/24 in its simplest form is 1/3.

3 0

2 years ago

Find the area of an isosceles trapezoid, if the lengths of its bases are 16 cm, and 30 cm, and the diagonals are perpendicular t

Klio2033 [76]

For this case we have:
a = 30 cm
c = 16 cm
We look for the length of the diagonal:
d = x + y
Where,
For x:
a ^ 2 = x ^ 2 + x ^ 2
x = a / root (2) = 30 / root (2) = 21.2132 cm
For y:
c ^ 2 = y ^ 2 + y ^ 2
y = c / root (2) = 16 / root (2) = 11.3137 cm
The diagonal is:
d = x + y
d = 21.2132 + 11.3137
d = 32.5269 cm
Then, the height is:
h = h1 + h2
For h1:
h1 = root (x ^ 2 - (a / 2) ^ 2) = root ((21.2132) ^ 2 - (30/2) ^ 2)
h1 = 15 cm
For h2:
h2 = root (y ^ 2 - (c / 2) ^ 2) = root ((11.3137) ^ 2 - (16/2) ^ 2)
h2 = 8 cm
Finally:
h = h1 + h2
h = 15 + 8
h = 23 cm
Then, the area is:
A = (1/2) * (a + c) * (h)
A = (1/2) * (30 + 16) * (23)
A = 529 cm ^ 2
Answer:
the area of an isosceles trapezoid is:
A = 529 cm ^ 2

4 0

2 years ago

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution o

DENIUS [597]

Answer:

33.36% probability that X is less than 2.

Step-by-step explanation:

The distribution is not normal, however, using the central limit theorem, it is going to be approximately normal. So

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

Normal Probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.2, \sigma = 1.4, n = 9, s = \frac{1.4}{\sqrt{9}} = 0.467$

(a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

This is the pvalue of Z when X = 2. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$