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3 years ago

Graph the linear equation y=×/3-4

Mathematics

1 answer:

shepuryov [24]3 years ago

5 0

I think so that graph will help you...................................

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Which of the following is the equation of the one that passes through the point

Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

the answer is C

( -6 , 1 )

y = -1/2x - 2

if x = -6

y = -1/2(-6) - 2

y = 3 - 2

y = 1

7 0

2 years ago

The following dataset represents the math test scores for a class of 20 students. 90, 85, 95, 100, 100, 90, 100, 65, 100, 85, 80

neonofarm [45]

Answer:

yes

Step-by-step explanation:

The mode is the measure of the central tendency for the given data set. The mode represents the highest frequency of the number

Since in the given data set as we can see that the 100 would be appeared 6 times

So this represent that the mode is 100

So here the mode would be the good measure

3 0

3 years ago

Alex scored 98,72, and 87 on his first three math test. What must he score on the next test to have an average of at least 86 (P

ss7ja [257]

Answer:

$\huge\boxed{\text{At least an } 87\%}$

Step-by-step explanation:

In order to find what score Alex must earn to have an average of 86 on all his tests, we need to first note what the formula to find the average of a data set is.

$\displaystyle \frac{x_1+x_2+...x_n}{n}$

What the formula means is that we have to add up all the values then divide by the total number of values.

Let's represent our unknown number as $x$ .

$\displaystyle \frac{98+72+87+x}{4} \geq 86$ (since we want AT LEAST an 86).

Let's solve this inequality for x.

$\frac{98+72+87+x}{4} \cdot 4 \geq 86 \cdot 4$
$98+72+87+x \geq 344$
$257 + x \geq 344$
$257 + x - 257 \geq 344 - 257$
$x \geq 87$

So, Alex must score at least an 87% on his next quiz to have an average of 86%.

Hope this helped!

8 0

2 years ago

Solve this equation v-10=-9

andrew11 [14]

First add 10 on both sides. So it will be v = 1.

6 0

3 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$