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3 years ago

What is the length of the line segment that has endpoints at (4, 7) and (-3, 9)?

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

6 0

Answer:

$\sqrt{53}$ or ≈7.280

Step-by-step explanation:

$l=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(4+3)^2+(7-9)^2}=\sqrt{53}$

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Olivia was pocking up some of her old stuff into a box. A box can hold 8

Citrus2011 [14]

Answer:

Step-by-step explanation:

2 easy ways:

1. Divide 7 by 3 then multiply by 2

7/3= seven thirds or 2.333

Now multiply by two to get 14 thirds

or 4.666

2. Multiply 7 by 2/3, or 0.666

This way is a lot easier with a

calculator

The weight of the box is 14/3’s or 4.666

6 0

3 years ago

Fill in the blank. In a drive thru performance study, the average service time for McDonald's is 203.21 seconds with a standard

Alenkasestr [34]

Answer:

5) 203.35

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The average service time for McDonald's is 203.21 seconds with a standard deviation of 5.67 seconds.

This means that $\mu = 203.21, \sigma = 5.67$

Sample of 90

This means that $n = 90, s = \frac{5.67}{\sqrt{90}} = 0.59767$

There is a 51% chance that the average drive-thru service time is less than ________ seconds.

X when Z is in the 51st percentile, that is, has a p-value of 0.51, so X when Z = 0.025.

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

By the Central Limit Theorem

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

$0.025 = \frac{X - 203.21}{0.59767}$

$X - 203.32 = 0.025*0.59767$

$X = 203.35$

203.35 seconds.

3 0

3 years ago

Can some one help me answer all ten questions they are due at 10:00 I will give out brainly. Part 2

Jlenok [28]

Answer:

solve for x

4x-6=26

4x=26+6

x=32/4

x=8

area of rectangle=l*b

=26*14

=364

area of triangle=b*h/2

=18.4*20.4/2

=375.36/2

=187.68 m^2

for another side of a triangle

(18.4)^2+(20.4)^2=c^2

338.56+416.16=c^2

$\sqrt{754.72}$ =c

27.47=c

27.5=c

now, perimeter of triangle=a+b+c

=27.5+18.4+20.4

=66.3 m^2

radius=diameter/2

=26/2

=13 ft

diameter=26 ft

circumference=2πr

=2*3.14*13

=81.64

Area of a circle=π $r^{2}$

=3.14*(13)^2

=3.14*169

=530.66 ft^2

Step-by-step explanation:

6 0

3 years ago

Simplify to create an equivalent expression

Keith_Richards [23]

B 33k- 15 hope it helps

5 0

3 years ago

When rounded to the nearest thousands an elephant weight is 5,000 pounds. What is the least amount the elephant could weigh? I n

ankoles [38]

Answer:

Step-by-step explanation:

It depends on what you have been told about rounding when a 5 is involved.

If you can round if the answer is 4500, then that would be your answer. If you have some complicated method of dealing with the 5 (involving odd or even numbers) the elephant's minimum weight will be different.

I would go with 4500.

7 0

3 years ago

Read 2 more answers