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3 years ago

GCF of 12z3, 45z2, and 18z5.

Mathematics

1 answer:

bija089 [108]3 years ago

4 0

Answer:

3z^2

Step-by-step explanation:

Factors of the different terms are ...

12z^3 = 2·2·3·z·z·z

45z^2 = 3·3·5·z·z

18z^5 = 2·3·3·z·z·z·z·z

The common factors are 3 and z·z.

The greatest common factor is 3z^2.

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What is 25% off of $34.99?

lesya [120]

25% is a fourth of something so first round $34.99 to about $35 and just get one fouth of it which would be a little less than $9 but if you put it into the calculator it will tell you that it is about $8.75 (to calculate using the calculator you need to chane the percentage to a decimal and multiply it)

6 0

3 years ago

The 3rd and 6th term of a geometric progression are 9/2 and 243/16 respectively find the first term, common ratio, seventh term

sashaice [31]

Answer:

Hello,

Step-by-step explanation:

$Let\ (u_n)\ the\ geometric\ progression.\\\\r\ is\ the\ common\ ratio.\\\\u_3=u_0*r^3\\u_6=u_0*r^6\\\\\dfrac{u_6}{u_3} =r^3=\dfrac{\frac{243}{16} }{\frac{9}{2} } =\dfrac{27}{8} =(\frac{3}{2} )^3\\\\\boxed{r=\dfrac{3}{2} }\\\\\\u_3=u_1*r^2 \Longrightarrow\ u_1=\dfrac{u_3}{r^2} =\dfrac{\frac{9}{2} }{(\frac{3}{2^2}) } =2\\\\\\u_7=u_6*\dfrac{3}{2} =\dfrac{729}{32}$

5 0

3 years ago

2 planes leave the airport at the same time. their speeds are 140 mph and 170 mph and the angle between their courses is 25 degr

Damm [24]

Answer:

183.27miles

Step-by-step explanation:

Note that we are considering two planes.

Plane 1

Speed = 140 mph

Time = 2.5 hours

Distance = ?

Speed = distance/time

140 mph = distance/2.5hrs

Distance = 140×2.5

= 350miles

Plane 2

Speed = 170 mph

Time = 2.5 hours

Distance = ?

Speed = distance/time

170 mph = distance/2.5hrs

Distance = 170×2.5

= 425miles

The angle between side is 25 degree

From cosine rule

c^2 = a^2 + b^2 − 2abcos(tetha)

Where a = 350

b = 425miles

c^2 = 350^2+425^2-2(350×425)(cos25)

c^2 = 122500+180625-297500(0.906)

c^2 = 303125-269535

c^2 = 33590

C = √33590

C = 183.27miles

The distance they flew apart after 2.5hrs is 183.27miles

5 0

2 years ago

In 2018, Mike Krzyewski and John Calipari topped the list of highest paid college basketball coaches (Sports Illustrated website

expeople1 [14]

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

$\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42$

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

$s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772$

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is $n = 10$

The significance level is $\alpha = 1 - 0.05 = 0.95$

The estimate, which is the sample standard deviation, is of $s = 0.8772$ .

Now, we have to find the critical values for the Pearson distribution. They are:

$\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228$

$\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004$

The confidence interval for the population variance is:

$\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}$

$\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}$

$0.3641 < \sigma^2 < 2.5646$

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

$\sqrt{0.3641} = 0.6034$

$\sqrt{2.5646} = 1.6014$

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

4 0

2 years ago

The measure of minor arc JL is 60°.

lara31 [8.8K]

We have been given an image of a circle. We are asked to find the measure of angle JKL.

We can see that angle JML is a central angle of our given circle, so measure of angle JML will be equal to measure of arc JL. So measure of angle JML will be 60 degrees.

We know that tangent of a circle is perpendicular to radius.

We can see that tangent KJ is tangent to circle at point J and tangent KL is tangent to circle at point L. This means that measure of angle KJM and angle KLM will be 90 degrees each.

We can see that JKLM is a quadrilateral. We know that all angles of a quadrilateral add up-to 360 degrees, so we can set an equation as:

$\angle JKL+\angle KLM+\angle LMJ+\angle MJK=360^{\circ}$