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3 years ago

Someone help please this was due a week ago

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

Yes , he or she is right .

katrin2010 [14]3 years ago

3 0

The answer is 120 inches

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Please help! Greatly appreciated sorry for really bad quality

Degger [83]

The first answer.

Point P is at (50,-40) the distance from Q to P is 80 units which you can find by subtracting the x of Q (-30) from the x of P (50).
50-(-30)=80
It then tells you point R is vertically above point Q so you know your x value for R will be the same as Q.
Add your distance from Q to P of 80 units to the y value of Q because you are traveling up.
-40+80=40
R will have a point of (-30,40) and a distance of 80 units

7 0

3 years ago

A car has a windshield wiper on the driver's side that has total arm length of 10 inches. It rotates

son4ous [18]

Answer:

75.44 Square Inches

Step-by-step explanation:

The diagram of the problem is produced and attached.

To determine the area of the cleaned sector:

Let the radius of the larger sector be R

Let the radius of the smaller sector be r

Area of the larger sector $=\frac{\theta}{360}X\pi R^2$

Area of the smaller sector $=\frac{\theta}{360}X\pi r^2$

Area of shaded part =Area of the larger sector-Area of the smaller sector

$=\frac{\theta}{360}X\pi R^2-\frac{\theta}{360}X\pi r^2\\=\frac{\theta \pi}{360}X (R^2- r^2)$

From the diagram, R=10 Inch, r=10-7=3 Inch, $\theta=95^\circ$

Therefore, Area of the sector cleaned

$=\frac{95 \pi}{360}X (10^2- 3^2)\\=75.44$ Square Inches$

7 0

3 years ago

Suppose you can somehow choose two people at random who took the SAT in 2014. A reminder that scores were Normally distributed w

Sindrei [870]

Answer:

22.29% probability that both of them scored above a 1520

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 1497, \sigma = 322$

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

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

$Z = \frac{1520 - 1497}{322}$

$Z = 0.07$

$Z = 0.07$ has a pvalue of 0.5279

1 - 0.5279 = 0.4721

Each students has a 0.4721 probability of scoring above 1520.

What is the probability that both of them scored above a 1520?

Each students has a 0.4721 probability of scoring above 1520. So

$P = 0.4721*0.4721 = 0.2229$

22.29% probability that both of them scored above a 1520

8 0

3 years ago

ANSWER FOR BRAINLEST (BEST TO EXPLAIN WILL GET IT)

lina2011 [118]

Answer:

-7g+13

Step-by-step explanation:

b) because combining like terms results in:

g-8g=7g

and

15-2= 13

5 0

2 years ago

2. The robot was initially placed at position (6,9), and at = 2 seconds, its position is (10,15).

qaws [65]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 6 &,& 9~) 
% (c,d)
&&(~ 10 &,& 15~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(10-6)^2+(15-9)^2}\implies d=\sqrt{4^2+6^2}\implies d=\sqrt{52}
\\\\\\
d\approx 7.2$

what's its speed? well, it traveled √(52) in 2 seconds, so is simply √(52)/2, which is about 3.6 units per second.

4 0

3 years ago