What will the 8th term in this pattern?:<br>400, 385, 370, 355, 340

What number is 25% of 7/8?

Oksanka [162]

0.25*7/8 = 1/4*7/8 = 7/32

3 0

3 years ago

Read 2 more answers

Which expression belongs

Nutka1998 [239]

For the expression to be equal to the original one, we have;

[(x + 1) * 5(x - 1)(x + 4)]/[(x - 1) * 7x]

<h3>How to Simplify Algebraic Expressions?</h3>

We are given the algebraic expression;

(5x² + 25x + 20)/(7x)

Now, looking at the numerator, a common factor to all terms is 5. Thus, we will factorize it out to get;

5(x² + 5x + 4) = 5((x + 1)(x + 4))

Now, we see that the expression that simplifies the algebra is given as;

[(x² + 2x + 1) * ( )]/[( ) * (7x² + 7x)]

Now, the numerator and denominator can be factorized to get;

[(x + 1)(x + 1) * ( )]/[( ) * 7x(x + 1)]

Thus, x + 1 will cancel out to get;

[(x + 1) * ( )]/[( ) * 7x]

For the expression to be equal to the original one, we have;

[(x + 1) * 5(x - 1)(x + 4)]/[(x - 1) * 7x]

Read more about Algebraic Expressions at; brainly.com/question/723406

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5 0

2 years ago

Ello, please help I’m a little stuck on how to go about solving this one. An explication would be great

Brut [27]

B, D and F are midpoints
so
DB = 1/2 (AE)
BF = 1/2 (CE)
DF = 1/2(AC)
in this case you only need to find out DF = ?
and you also know AC = 50
so
DF = 1/2 (AC)
DF = 1/2(50)
DF = 25
answer
DF = 25

8 0

3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 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 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.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

4 years ago

Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1,3).

Nataliya [291]

Answer:

Step-by-step explanation:

for two lines to be parallel they MUST have the same slope but have different Y intercepts ( Y axis crossing values)

parallel lines will never intersect each other

first I use a graphing calcuator to solve the problem

I like to use y intercept form better, but you don't have use it

I will try to solve it directly without going through the y intercept equation

x + 5y = 10 in y intercept form y = mx + b is

5y = -x + 10 divide both sides by 5

y = -x/5 + 10/5

y = -x/5 + 2

Now I need to find a parallel line that passes through the x =1 and y = 3 point recall this line will have the same slope = m = -1/5

and a different Y axis crossing point 'b' that we don't know

y = -x/5 + b y = 3 when x = 1 so slove for 'b'

3 = -1/5 + b

3 = -0.2 + b add -0.2 to both sides

3.2 = b

y = -x/5 + 3.2 this is answer in y intercept form

if you multiply both sides by 5

5y = -x + 16 add x to both sides

x + 5y = 16 answer in standard form

THE EASIER WAY TO SOLVE THE PROBLEM IS ......

x + 5y = 10 find a line parallel that passes through x = 1, y = 3

x + 5y = Z Z is an unknown value plug in x and y and solve for Z

1 + 5(3) = Z

Z = 16 so the parallel line in standard form is

x + 5y = 16

same as before, my y intercept method was correct, but not worth the effort

6 0

3 years ago

What will the 8th term in this pattern?:400, 385, 370, 355, 340