Are these angles supplementary or congruent,

Reduce 36/45 to lowest terms and write the denominator in the blank.

Firlakuza [10]

Answer:

4/5

Step-by-step explanation:

Step 1:

36/45 = 12/15

Step 2:

12/15 = 4/5

Answer:

4/5

Hope This Helps :)

5 0

3 years ago

Read 2 more answers

What is f(0) if f(t) = t^2 – 2t – 2?

Dafna11 [192]

Answer:

-2

Step-by-step explanation:

f(t) = t^2 – 2t – 2

To find f(0), replace t with 0 in the expression and evaluate the expression.

f(0) = 0^2 – 2(0) – 2

f(0) = -2

6 0

2 years ago

Read 2 more answers

Which are solutions of 1 < 3x – 2 < 13?

AURORKA [14]

Answer:

0.076 or 4.409 try both its one of those two

Step-by-step explanation:

6 0

1 year ago

4x - 5y = 15<br> find the slope and y-intercept

Mazyrski [523]

Answer:

slope=4/5

y-intercept = -3

Step-by-step explanation:

first we need to rewrite the expression to

$y=mx+b$

were b is the y-intercept and m the slope

$4x-5y=15\\\\4x-5y+5y=15+5y\\\\4x+0=15+5y\\\\4x=5(3)+5(y)\\\\4x=5(3+y)\\\\\frac{1}{5}*4x= \frac{1}{5}*5(3+y) \\\\\frac{4x}{5}=3+y\\\\\frac{4x}{5} -3=3+y-3\\\\\frac{4x}{5} -3=y+0\\\\y=\frac{4x}{5} -3\\\\y=\frac{4}{5}x-3$

so there we have it

$m=\frac{4}{5} \\\\b=-3$

6 0

2 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$