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Amanda [17]
3 years ago
15

What are the missing statements in steps 6 and 10 of the proof? Statement 6: Statement 10: Statement 6: || Statement 10: || Stat

ement 6: Statement 10: AEB BEC Statement 6: Statement 10: Statement 6: ABC BCD Statement 10: BAD ADC NextReset
Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Step-by-step explanation:

leva [86]3 years ago
5 0
I have the same question I need help with
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(r/s)(3) find the value.
klasskru [66]
3r
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4 0
3 years ago
Please help thanks very much
Shalnov [3]
Angle=1/2(big arc-small arc)
x=1/2(360-x)-x
2x=360-2x
4x=360
x=360/4=90
Answer 90⁰
7 0
3 years ago
The slope of the the tangent line to a curve at a point. Calc derivative help
MrRissso [65]

Hello from MrBillDoesMath!

Discussion:

The slope of the tangent to the curve y = 4x^3   is given by the first derivative.

-----------------------------------

y' = 4 (3x^2) =  12x^2

At (-3,-108), the slope is   12x^2 = 12(-3)^2 = 12*9 = 108

-----------------------------------

-----------------------------------

y = mx + b

From the first part, m = 108 so y = 108x + b.

Substituting  y = -108 when x = -3 gives

-108 =  108(-3) + b =>   add 3* 108 to both sides

-108 + 108(3) = b   =>   as 3 -1  = 2

108*2 = b = 216

y = 108x + 216

-----------------------------------

Thank you,

MrB

5 0
3 years ago
Margarite owns a boutique. She sells her jewelry at a 30% markup. If she buys a bracelet for $22, what would she sell it for aft
olga nikolaevna [1]

Answer:

selling price is $26.4

Step-by-step explanation:

Here, we want to apply the markup percentage so as to get the price of the jewelry

mathematically, we have the selling price as;

$22 + (30% of $22)

= $22 + 4.4

= $26.4

4 0
3 years ago
Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.
blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean \mu=510,

Standard deviation \sigma=100

Sample mean x=300

Percentage of students scored below 300 is given by,

P(Z\leq \frac{x-\mu}{\sigma})\times 100

=P(Z\leq \frac{300-510}{100})\times 100

=P(Z\leq \frac{-210}{100})\times 100

=P(Z\leq-2.1)\times 100

=0.0179\times 100

=1.79\%

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

P(x\leq t)=0.90

Now, P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90

P(Z< \frac{t-\mu}{\sigma})=0.90

\frac{t-\mu}{\sigma}=1.28

\frac{t-510}{100}=1.28

t-510=128

t=128+510

t=638

Score puts someone in the 90th percentile is 638.

5 0
3 years ago
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