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2 years ago

12

If QR=x+29 and PS=x–16, what is the value of x?

1 answer:

Vinil7 [7]2 years ago

3 0

Answer:

61

Step-by-step explanation:

By the triangle midsegment theorem, 2(SP)=RQ.

So, 2(x-16)=x+29, meaning x=61.

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The cheetah is 7/6 the Perigen falcon is 10/3

Brilliant_brown [7]

The answer is $2 \frac{1}{6}$ .

Explanation:

In order to subtract the fractions, we must make them like fractions. To do this, the denominators must be the same by multiplying (only). Since the first fraction is $\frac{10}{3}$ , 3 can be multiplied by 2 to get 6, which is the other fraction, we can multiply it. Whatever you do to the denominator you must do to the numerator. Now multiply 2 by the numerator (10) to get $\frac{20}{6}$ . Now we can subtract the fractions $\frac{20}{6}$ and $\frac{7}{6}$ to get 13/6. Since this fraction is not in mixed fraction form yet, we must do that first. goes into 13 twice, so the whole number is 2 and there is still 1 left, making the fraction $\frac{1}{6}$ . Therefore, the difference is 2 $\frac{1}{6}$ .

8 0

3 years ago

The diameter of the base of the cone measure 17units the height measures 15units what is the unfilled volume inside the cylinder

777dan777 [17]

Answer:

V = 1134.9

Step-by-step explanation:

Since $V=\pi r^{2}\frac{h}{3}$

Firstly we need to divide 17 by 2 to get the radius which is 8.5

Then we need to solve by putting the values in...

$=\pi$ x $8.5^{2}$ x $\frac{15}{3}$

= 1134.9

4 0

3 years ago

Please answer quickly!

blsea [12.9K]

Your answer is D. This is because a number multiplied by itself is x².

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2 years ago

An Xത chart with three-sigma limits has parameters as follows: UCL = 104; Center Line = 100; LCL = 96; n = 5 Suppose the process

ruslelena [56]

Answer:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

Step-by-step explanation:

We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.

The probability of one point of being over UCL=104 is:

$z=(X-\mu)/\sigma=(104-98)/8=6/8=0.75\\\\P(X>104)=P(z>0.75)=0.227$

The probability of one point of being under LCL=96 is:

$z=(X-\mu)/\sigma=(96-98)/8=-2/8=-0.25\\\\P(X$

Then, the probability of exhibit lack of control is:

$P=P(X>104)+P(X$

The probability of having at least one point out of control in the first three points is:

$P(x\geq1)=1-P(0)\\\\P(x\geq1)=1- 0.052=0.948\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{3}= 1*1*0.0515=0.0515\\\\$

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

7 0

2 years ago

Kenton completed 30 percent of a 90 page reading assignment how many pages did he read

Eddi Din [679]

90*0.30 = 27

i think if i am right it owuld be 27 pags

4 0

3 years ago

Read 2 more answers