Whats the greatest common factor of 25 and 100

Solve for x. 4(x + 3) = x +42 x = [?]

laila [671]

Answer: x=10

Step-by-step explanation:

$4(x+3)=x+42\\4x+12=x+42\\4x-x=42-12\\3x=30\\x=10$

7 0

3 years ago

64÷54 is the number of.... whar

ankoles [38]

Your answer is 1.18518519

8 0

3 years ago

Read 2 more answers

Which statement is true about the given information?

Lubov Fominskaja [6]

Answer

∴ The true statement is Answer:

The true statement is BD ≅ CE ⇒ 3rd answer

Step-by-step explanation:

- There is a line contained points B , C , D , E

- All points are equal distance from each other

- That means the distance of BC equal the distance of CD and equal

the distance of DE

∴ BC = CD = DE

- That means the line id divided into 3 equal parts, each part is one

third the line

∴ BC = 1/3 BE

∴ CD = 1/3 BE

∴ DE = 1/3 BE

∵ BC = CD

∴ C is the mid-point of BD

∴ BC = 1/2 BD

∵ CD = DE

∴ D is the mid-point of DE

∴ CD = 1/2 CE

- Lets check the answers

* BD = one half BC is not true because BC = one half BD

* BC = one half BE is not true because BC = one third BE

* BD ≅ CE is true because

BD = BC + CD

CE = CD + DE

BC ≅ DE and CD is common

then BD ≅ CE

* BC ≅ BD is not true because BC is one half BD

∴ The true statement is BD ≅ CE

plz mark me brainliest

5 0

2 years ago

Here is the question !

Delvig [45]

Answer:

Step-by-step explanation:

Co-ordinates of F, G and H are:

F (-4, 5)

G (1, 10)

H (-9, 10)

You can find the lengths of sides by using the distance formula:

length(FG) = $\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ = $\sqrt{(1-(-4))^{2} + (10-5)^{2}} = 5\sqrt{2}$ units

Similarly,

length(GH) = 10 units

length(HF) = $5\sqrt{2}$ units

Hence, perimeter will be:

P = length(FG) + length(GH) + length(HF) = $10(\sqrt{2} + 1)$ units

3 0

3 years ago

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 $\alpha = 0.95$

So

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

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 $\alpha = 0.99$

So

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

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.

4 0

3 years ago