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3 years ago

PLEASE HELP! ITS GEOMETRY!! THE PICTURE IS PROVIDED BELOW PLZ HELP..... (┬┬﹏┬┬)

Mathematics

1 answer:

777dan777 [17]3 years ago

3 0

Answer:

hello I really do not get how yo read it because the words are so small sorry

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Miguel spent 1/8 of his savings on a new video game system. Mila spent 12% of her savings on her DVD, and Manny spent 0.1 of his

Vilka [71]

Miguel spent the most of his savings.

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3 years ago

I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an

Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

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.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

$\mu = 0 - 5 = -5$

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

$\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}$

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

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

$Z = \frac{1 - (-5)}{\sqrt{13}}$

$Z = \frac{6}{\sqrt{13}}$

$Z = 1.66$

$Z = 1.66$ has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

5 0

3 years ago

Start a conversation/argument in the comments

Digiron [165]

Answer:

What do we start a convo or an argumnet about...and why?

Step-by-step explanation:

but heres one : Donald Trump or Biden?

heheheheheheh

5 0

3 years ago

Read 2 more answers

To make 1 batch of vegetable soup, Sam needs 16 ounces of water and 8 ounces of vegetables. How does the amount of water compare

BARSIC [14]

The amount of water is double in comparison to the amount of vegetables.

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3 years ago

#10 i The table shows the admission costs (in dollars) and the average number of daily visitors at an amusement park each the pa

lions [1.4K]

The line of best fit is a straight line that can be used to predict the

average daily attendance for a given admission cost.

Correct responses:

The equation of best fit is; $\underline{ \hat Y = 1,042 - 4.9 \cdot X_i}$

The correlation coefficient is; r ≈<u> -0.969</u>

<h3>Methods by which the line of best fit is found</h3>

The given data is presented in the following tabular format;

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|}Cost, (dollars), x&20&21&22&24&25&27&28&30\\Daily attendance, y&940&935&940&925&920&905&910&890\end{array}\right]$

The equation of the line of best fit is given by the regression line

equation as follows;

$\hat Y = \mathbf{b_0 + b_1 \cdot X_i}$

Where;

$\hat Y$ = Predicted value of the<em> i</em>th observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

$X_i$ = The <em>i</em>th observed value

$b_1 = \mathbf{\dfrac{\sum (X - \overline X) \cdot (Y - \overline Y) }{\sum \left(X - \overline X \right)^2}}$

$\overline X$ = 24.625

$\overline Y$ = 960.625

$\mathbf{\sum(X - \overline X) \cdot (Y - \overline Y)} = -433.125$

$\mathbf{\sum(X - \overline X)^2} = 87.875$

Therefore;

$b_1 = \mathbf{\dfrac{-433.125}{87.875}} \approx -4.9289$

Therefore;

The slope given to the nearest tenth is b₁ ≈ -4.9

$b_0 = \mathbf{\dfrac{\left(\sum Y \right) \cdot \left(\sum X^2 \right) - \left(\sum X \right) \cdot \left(\sum X \cdot Y\right)} {n \cdot \left(\sum X^2\right) - \left(\sum X \right)^2}}$

By using MS Excel, we have;

n = 8

∑X = 197

∑Y = 7365

∑X² = 4939

∑Y² = 6782675

∑X·Y = 180930

(∑X)² = 38809

Therefore;

$b_0 = \dfrac{7365 \times 4939-197 \times 180930}{8 \times 4939 - 38809} \approx \mathbf{1041.9986}$

The y-intercept given to the nearest tenth is b₀ ≈ 1,042

The equation of the line of best fit is therefore;

$\underline{\hat Y = 1042 - 4.9 \cdot X_i}$

The correlation coefficient is given by the formula;

$\displaystyle r = \mathbf{\dfrac{\sum \left(X_i - \overline X) \cdot \left(Y - \overline Y \right)}{ \sqrt{\sum \left(X_i - \overline X \right)^2 \cdot \sum \left(Y_i - \overline Y \right)^2} }}$