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allsm [11]
2 years ago
5

Pls solve I need it solved rn show work urgent PLS*****

Mathematics
1 answer:
omeli [17]2 years ago
5 0

Answer:

16. y=-5x-20; 17. y=6x+18, or factored y=6(x+3)

Step-by-step explanation:

Your slope is -5

y=-5x+b, solve for b by pluging the point (-3,-5) in for x and y

-5=-5(-3)+b, solve for b

-20=b, now rewrite the equation

y=-5x-20

Your slope is 6

y=-5x+b, solve for b by pluging the point (-3,-0) in for x and y

0=6(-3)+b, solve for b

18=b, now rewrite the equation

y=6x+18, or factored y=6(x+3)

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8 0
3 years ago
the hypotenuse of a 30-60-90 degree right triangle has a measure of 10. what is the measure of the shortest side
VARVARA [1.3K]

Answer:

5

Step-by-step explanation:

A 30-60-90 triangle's sides equal as follows:

Hypotanuse= x (a number)

Long side= x/2( also can be calculated as x* 1/2) * \sqrt{3}

Short side = x/2 or x * 1/2

5 0
2 years ago
С E A B G 77° D plz help.me​
Snowcat [4.5K]
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2 years ago
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
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Answer:

240 to 60 or in simpler form 1 and 4

Step-by-step explanation:

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