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3 years ago

8

A car enters a highway driving 60 mph. A truck enters the highway 6 minutes later and drives 65 mph. The vehicles travel a total

of 56 miles and arrive at the same time. How far did the car travel?

1 answer:

algol [13]3 years ago

4 0

Answer:

65 or 60

Step-by-step explanation:

I am not really positive about my answers so please take this into consideration.

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Determine f(-a) where f(x) = x2 - 6x and simplify.

jeka57 [31]

Answer:

6a

Step-by-step explanation:

replace x with -a

-2a -6(-a)

-2a +8a

6a

5 0

2 years ago

Write an expression for the situation

MAXImum [283]

Answer:

3b

Step-by-step explanation:

if each brick is 3 inches tall, and there are b bricks, each b is 3, therefore 3b

8 0

3 years ago

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

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

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.

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 $\frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33$

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

$2.33 = \frac{X - 117}{4.33}$

$X - 117 = 2.33*4.33$

$X = 127.1$

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0

4 years ago

Solve the equation below

Agata [3.3K]

Set up the equation:

4(8.75x-2)=97
35x-8 =97
35x =97+8
35x =105
x. =3

The value of x is 3.

7 0

3 years ago

Infinite over E n=1 4(0.5)^n-1

yan [13]

Answer:

S = 8

Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

$\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}$

We can also write in summation notation rather S-term as:

$\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}$

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

$\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}$

Therefore, the sum will converge to 8.

Please let me know if you have any questions!

6 0

2 years ago