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3 years ago

What is the solution to the equation below? Vx + 3 = x - 3 A. 3 B. 6 C. 5 D. 4

Mathematics

1 answer:

kherson [118]3 years ago

6 0

Answer:

Step 1: Solve Square Root

Vx+3=x-3

x+3=(x-3)^2 (squared both sides)

x+3=x^2-6x+9

x+3-(x^2-6x+9)=0

(-x+1)(x-6)=0 (factor left side of equation)

-x+1=0 or x-6=0

x=1 or x=6

When you plug it in to check

1 (Doesn't Work)

6 (Work)

Therefore, 6 is your solution.

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Find the center and radius of the circle whose diameter has an endpoint at (-3, -4) and the origin.

Zina [86]

Given end points of diameter,

(x1,y1)=(-3,-4)

(x2,y2)=(0,0)

Now,

the equation of circle is,

(x-x1)(x-x2)+(y-y1)(y-y2)=0

or, (x+3)(x-0)+(y+4)(y-0)=0

or, x^2 +3x +y^2 +4y =0

or, x^2 +y^2 +3x + 4y=0

which is in the form of x^2 +y^2 +2gx +2fy + c=0

where,

g=3/2

h=2

c=0

Now,

$radius(r) = \sqrt{ {g}^{2} + {f}^{2} - c } \\ = \sqrt{ \frac{ {3}^{2} }{ {2}^{2} } + {2}^{2} - 0 } \\ = \sqrt{ \frac{9}{4} + 4} \\ = \sqrt{ \frac{25}{4} } \\ = \frac{5}{2}$

6 0

3 years ago

The volume of 10

Alona [7]

Hello!

1 floz divided by 100 = 0.01
0.01 x 10 = 0.1 floz for 10 drops.

8 0

3 years ago

Read 2 more answers

What is the answer help ASAP!! :))

vampirchik [111]

Answer:

Step-by-step explanation:

8 0

3 years ago

A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once p

____ [38]

Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

step 1:-

Given sample size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

Population proportion of employed individuals who work at home at least once per week P = $\frac{x}{n} =\frac{41}{200} =0.205$

Q=1-P= 1-0.205 = 0.705

step 2:-

Now $\sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }$

=0.0015

step 3:-

Confidence intervals

using formula

$(P - Z_∝} \sqrt{\frac{P Q}{n},} (P + Z_∝} \sqrt{\frac{P Q}{n},$

$(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887$

=0.20113,0.20887[/tex]

conclusion:-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

4 0

3 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago