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3 years ago

title=" \frac{3}{5} {x}^{2} - 2 = - 5" alt=" \frac{3}{5} {x}^{2} - 2 = - 5" align="absmiddle" class="latex-formula">
single variable quadratics

Mathematics

1 answer:

stira [4]3 years ago

7 0

Firstly, add 2 to both sides of the equation: $\frac{3}{5}x^2=-3$

Next, multiply both sides by 5/3: $x^2=-5$

Next, square root both sides: $x= \pm \sqrt{-5}$

Lastly, factor out i (i = √-1) and <u>your answer will be $x= \pm \sqrt{5}i$ </u>

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X<9.7 on a line graph

fiasKO [112]

The inequality x < 9.7 tells us x needs to smaller than 9.7.
A number like 5 will be fine. Because 5 < 9.
But someone like 10 won't be fine.
Because 9 < 10.

Even 9.7 won't be suffice.

Because 9.7 < 9.7 does not make sense. If it was 9.7 <= 9.7 then it would be fine.

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2 years ago

(x-1)(x-3)(x+5)(x+7)=297

mihalych1998 [28]

First simplify the expression into polynomial form,

$(x-1)(x-3)(x+5)(x+7)=297$

$x^4+8x^3-10x^2-104x+105=297$

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Now factor into,

$(x-4)(x+8)(x^2+4x+6)=0$

Which means the solutions are,

$x-4=0\implies\boxed{x_1=4}$

$x+8=0\implies\boxed{x_2=-8}$

and then two complex solutions because determinant of the third factor $D\lt0$ ,

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$x^2+4x+4=-2$

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2 years ago

Read 2 more answers

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

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2 years ago

3 groups of 5 how<br> Many are there in total

Vadim26 [7]

It's 15 because

3 groups and 5 people in each group

and becomes 15

I HOPE THIS HELPS

7 0

2 years ago

Answer questions 1, 2, 3 and challenge

hjlf

1. find volume
V=LWH
V=25*13*14
V=4550
each covers 1000
how many will cover the whole thing?
we can't put 4 becuase that leaves 550 uncovered
add another

5 is answer (it is too much , but at leas not too little) answer is D

2.
V=LWH
V=6*4*10
V=240

answer is D

3. count how many are there, don't forget the hidden ones
13 cubes
volume=legnth tiems width times height
cubes so
v=side^3
side=3
3^3=27
13 cubes
13*27=351 in^3

challenge (this is meant to challenge you, not for you to ask people, but I will solve anyway, robbin you of experience)

V=LWH
V=5880
H=30
5880=LW30
divide both sides by 30
196=LW
they have square bases so L=W
196=L^2
sqrt both sides
14=L=W
the side legnth is 14in

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3 years ago