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3 years ago

I don't understand 14. A, B, and C

Mathematics

1 answer:

Luden [163]3 years ago

7 0

14a. C = 0.09(m - 25) + 50
C = 0.09(80 - 25) + 50
C = 0.09(55) + 50
C = 4.95 + 50
C = 54.95

14b. C = 0.09(m - 25) + 50
85 = 0.09(m - 25) + 50
- 50 - 50
35 = 0.09(m - 25)
0.09 0.09
388⁸/₉ = m - 25
+ 25 + 25
413⁸/₉ = m

14c. The input values that are below 25 does make any sense because it shows that the input values can make the output values decrease itself.

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HELP ME PLEASE!!! MATHEMATICS

sukhopar [10]

By Pythagoras AB^2 = 12^2 - 6^2 = 108

AB = sqrt 108 = 10.39 to nearest hundredth

The perpendicular from M to DC will be parallel and equal to AB so it = 10.39.

Also AB^2 = CB * DB

108 = 6 * DB
DB = 108/6 = 18
so DC = 18-6 = 12

MD^2 = (1/2*12)^2 + 108 = 144

so MD = 12

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4 years ago

Write an equation for the relationship between x and y.

wariber [46]

The first one Y=3x:)

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3 years ago

Find r if 0=pi/6 rad sector 64m^2

kap26 [50]

Given:

Area of a sector = 64 m²

The central angle is $\theta=\dfrac{\pi}{6}$ .

To find:

The radius or the value of r.

Solution:

Area of a sector is:

$A=\dfrac{1}{2}r^2\theta$

Where, r is the radius of the circle and $\theta$ is the central angle of the sector in radian.

Putting $A=64,\theta=\dfrac{\pi}{6}$ , we get

$64=\dfrac{1}{2}r^2\times \dfrac{\pi}{6}$

$64=\dfrac{\pi}{12}r^2$

$64\times \dfrac{12}{\pi}=r^2$

$\dfrac{768}{\pi}=r^2$

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$\sqrt{\dfrac{768}{\pi}}=r$

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3 years ago

The scores of fourth grade students on a mathematics achievement test follow a normal distribution with a mean of 75 and standar

mamaluj [8]

Answer:

The appropriate solution is:

(a) 0.1056

(b) 0

(c) 0.9544

Step-by-step explanation:

The given values are:

Mean,

$\mu = 75$

Standard deviation,

$\sigma = 4$

(a)

⇒ $P(x>80)=1-(x$

$=1-P[\frac{x-\mu}{\sigma}$

$=1-P(z$

By using the table, we get

$=1-0.8944$

$=0.1056$

(b)

According to the question, the values are:

$n$ = 64

$\mu_\bar{x}$ = 75

Now,

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

= $\frac{4}{\sqrt{64} }$

= $\frac{4}{8}$

= $0.5$

⇒ $P(\bar {x} >80 ) = 1 - P(\bar x$

$=1 - P[\frac{(\bar x-\mu_\bar x)}{\sigma \bar x} < \frac{80-75}{0.5} ]$

$=1-P(z$

By using the table, we get

$=1-1$

$=0$

(c)

As we know,

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

= $\frac{4}{\sqrt{64} }$

= $\frac{4}{8}$

= $0.5$

then,

= $P(74< \bar x$

= $P[\frac{74-75}{0.5} < \frac{\bar x-\mu \bar x}{\sigma \bar x} < \frac{76-75}{0.5} ]$

= $P(-2$

= $P(z$

By using the table, we get

= $0.9772-0.0228$

= $0.9544$

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3 years ago

Mehek14 i would love for you to be the one the help me other ppl u try 2 c:

devlian [24]

Question 1: X = 2/3
Question 2: X = 4
Question 3: X = 48
Question 4: ?
Question 5: X = 1
Question 6: Yes
Question 7: No
Question 8: No
Question 9: Kara (Danvers is always right) < ignore that
Question 10: X = 0.5
Hope this helped!

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3 years ago

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