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Neporo4naja [7]
2 years ago
14

HELP! WILL GIVE BRANLIEST!

Mathematics
2 answers:
Brrunno [24]2 years ago
8 0
6x+x-8=90 because right angle
7x=98 combine like terms
7x/7=98/7
X=14
m<1=6*14=84
m<2=14-8=6
laila [671]2 years ago
8 0

Answer:

ITS X=14 GURL TRUST

Step-by-step explanation:

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Draw and classify the polygon with the given vertices. Find the perimeter and the area of the polygon. M(-2,5), N(3,-2), P(-2,-2
Nadusha1986 [10]

Answer:

<em>Perimeter = 20.6cm  to 1dp  = 20.60cm to 2dp</em>

<em>Area = Base = 8.6/2 = 4.3 x 4.069  = 17.5cm^2</em>

<em>Height is found 4.07  cos( 54.47deg) 5/8.6 = 4.069cm</em>

Step-by-step explanation:

<em>Right angle side triangle, sides </em>

<em>= MN =8.6cm </em>

<em>= NP=5cm </em>

<em>= PM=7cm </em>

<em>= 25sq + 49sq = 74sq</em>

<em>MN^2 =√74 = 8.60232526704 = 8.6cm</em>

<em>P = 8.6 + 5 + 7 = 20.6cm </em>

5 0
3 years ago
Read 2 more answers
Rewrite `10^{32}\ \/ \ 10^{6}\ `using a single exponent.
IrinaVladis [17]

Answer:

10^38

Step-by-step explanation: :)

8 0
3 years ago
Help me pleaseeeeeee
Pepsi [2]

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5 0
2 years ago
HELP PLEASE!!!!!!!!?!??
lianna [129]

Answer:

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Step-by-step explanation:

I think this is correct

6 0
3 years ago
Read 2 more answers
The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0
3 years ago
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