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tiny-mole [99]
2 years ago
11

Point M is on line segment \overline{LN} LN . Given LN=17LN=17 and MN=3,MN=3, determine the length \overline{LM}. LM .

Mathematics
1 answer:
Alinara [238K]2 years ago
8 0

Answer:

14

Step-by-step explanation:

the full length of the line is 17, half of it is 3.

subtract line LN from MN to get LM.

17-3=14

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The coordinates of A, B,and C in the diagram are A(p,4), B(6,1), C(9,q). What equation correctly relates p and q?
murzikaleks [220]
I don't see any diagram. So, I'll just wing it.

Value of q:
p, 6, 9 ⇒ there is a difference of 3. The sequence increases by 3. So, it can be assumed that the p is equal to 3. A(p,4) ⇒ A(3,4)

Value of q:
4, 1, q ⇒ there is a difference of 3. The sequence decreases by 3. So, it can be assumed that q is equal to -2. C(9,-2)
 
        p        q
A     3        4     ⇒ 3 + 4 = 7
B     6        1     ⇒ 6 + 1 = 7
C     9       -2    ⇒ 9 + (-2) = 7

Notice that the sequence has an equation of p + q = 7.


7 0
3 years ago
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Please round to nearest hundredth and give explanation.
skelet666 [1.2K]

Answer:

I think PR = 20 but I'm not sure

8 0
2 years ago
Which of the following systems of inequalities has point D as a solution?<br> f(x) = 3x+4
Alex73 [517]

I think is this one
8 0
3 years ago
Does anyone know this
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6 0
2 years ago
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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
2 years ago
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