Can someone help me with this task please please !!!!

Solve the system of equations.

postnew [5]

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:

(20,−1)

Equation Form:

x=20,y=−1

plz mark me as brainliest :)

5 0

3 years ago

Jaden is looking up at the top of a 56 Foot

marshall27 [118]

noooooooooooooooooooolo

7 0

3 years ago

50 POINTS! IM SO CONFUSED I'LL MARK BRAINLIEST!!!!

yan [13]

The function of x is $\bold{\frac{-1x}{2}}$ .

<u>SOLUTION:</u>

Given that, we have to write a linear function f with function of (-4) =2 and function of 6=(−3).

Now, let the linear function be $\text{function of x}=a x+b$

Then, function of -4 = $a(-4)+b \rightarrow 2=-4 a+b \rightarrow b=2+4a \rightarrow (1)$

And, function of 6 = $a(6)+b \rightarrow-3=6 a+b \rightarrow b=-3-6a \rightarrow (2)$

On equating (1) and (2) to find the value of a,

$2+4a=-3-6a$

On grouping the common terms,

$4a+6a=-3-2 \rightarrow 10a=-5 \rightarrow a=\frac{-5}{10}$

$\Rightarrow a=\frac{-1}{2} \rightarrow (3)$

On substituting the value of (3) in (1) we get,

$b=2+4a \rightarrow b=2+4\times\frac{-1}{2} \rightarrow b=2+\frac{-4}{2} \rightarrow b=2+(-2)$

$\Rightarrow b=0$

So, the function of x = $\frac{-1x}{2}+0$

4 0

3 years ago

30-60-90 triangle..find the value of x

Zepler [3.9K]

Answer:

x=8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 71 and a stan

kumpel [21]

Answer:

The lowest score on the final exam that would qualify a student for an A is 80.

The lowest score on the final exam that would qualify a student for a B is 74.68.

The lowest score on the final exam that would qualify a student for a C is 67.33.

The lowest score on the final exam that would qualify a student for a D is 62.

Step-by-step explanation:

When the distribution is normal, we use 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.

Mean of 71 and a standard deviation of 7.

This means that $\mu = 71, \sigma = 7$

Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's.

This means that:

90th percentile and above: A

70th percentile and below 90th: B

30th percentile to the 70th percentile: C

10th percentile to the 30th: D

Lowest score for an A:

Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.

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

$1.28 = \frac{X - 71}{7}$