What is the solution to this system of linear equations? y − x = 6 y + x = −10

What is y - 2 = -3 [x-7] written in standard form

Akimi4 [234]

Y=-3x+23

distribute -3, get y alone (add 2 to both sides)

3 0

2 years ago

An expression is shown.

Vikki [24]

Answer:

Step-by-step explanation:

2+8+3_5x1

7 0

3 years ago

A woman is randomly selected from the 18-24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

dybincka [34]

Answer:

0.0274

Step-by-step explanation:

The mean is $\mu =114.8$ and the standard deviation is $\sigma =13.1.$

Calculate

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

for $X=140:$

$Z=\dfrac{140-114.8}{13.1}\approx 1.9237.$

If $X\sim N(114.8,\ 13.1),$ then $Z\sim N(0,1)$

and

$Pr(X>140)=Pr(Z>1.9237).$

Use table for normal distribution probabilities to get that

$Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.$

7 0

3 years ago

Read 2 more answers

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

If x and y vary directly and x=25 when y=100 find x when y=24. PLEASE HELP FASTTTTTTT

Veseljchak [2.6K]

Answer:

x = 6

Step-by-step explanation:

Given that x and y vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition x = 25 when y = 100 , then

100 = 25k ( divide both sides by 25 )

4 = k

y = 4x ← equation of variation

When y = 24 , then

24 = 4x ( divide both sides by 4 )

6 = x

6 0

2 years ago