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krok68 [10]
3 years ago
9

This won’t take long please look at it

Mathematics
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

I cant see the whole picture

Step-by-step explanation:

.

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I NEED HELP FAST PLEAASE I BEG U!!!!!
motikmotik

Answer:

You must mix 2.67 gals. of pure alcohol to obtain the desired mixture.

4 0
3 years ago
Ik rp because the i forgot the sc
Stels [109]

Answer:

Step-by-step explanation:

This is a geometric mean problem...a ratio. The geometric mean is the height of the triangle, 6. To solve:

\frac{x}{6} =\frac{6}{3} and cross multiply to solve for x:

3x = 36 so

x = 12

6 0
1 year ago
Read 2 more answers
One terameter equals 1012 meters. One micrometer equals 10 meter.
Butoxors [25]

Answer:

<h2>a)  10⁶metres</h2><h2>b) 10¹⁸metres</h2>

Step-by-step explanation:

Given One terameter equals 10¹² meters. One micrometer equals 10⁻⁶ meter, one nanometer equals 10⁻⁹meter;

a) The product of one terameter and one micrometer will be expressed as;

10¹²  * 10⁻⁶

Since the both have the same base, we will add their power according to one of the law in indices as shown

10¹²  * 10⁻⁶ = 10⁻⁶⁺¹²

= 10⁶metres

b) The quotient of one terameter and one micrometer can be gotten by taking their ratio as shown below;

Note that when taking the quotient, the power are subtracted

= \frac{10^{12} }{10^{-6} } \\= 10^{12-(-6)}  \\= 10^{12+6} \\= 10^{18}metres

3 0
3 years ago
1.) (-18, -20) (18, 16)
aleksandrvk [35]

1.) (-18, -20) (18, 16)

m= 1

2.) (-7, 14) (-11, 4)

m= 5/2

3.) (-5, -4) (3, 9)

m= 13/8

4.) (13, -8) (-9, -1)

m= -7/22

3 0
3 years ago
he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
3 years ago
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