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3 years ago

8

A window washer cleaned 38 windows in 2 hours. At this rate, how many windows did he clean in 7 hours?

2 answers:

andriy [413]3 years ago

6 0

Answer:

Step-by-step explanation:

38/2=19x7=133

Kryger [21]3 years ago

3 0

Answer:

I believe its 19 times 7 which is 133. I did this in my head so correct me if im wrong lol.

Step-by-step explanation:

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What is the equation of the linear function represented by the table?

Korolek [52]

Hello!

You can put x into the answer choices to see if we get y
----------------------------------------------------------------------------------------------------
y = -x + 9
y = -(-5) + 9
y = 5 + 9
y = 14

This could be a answer

y = -(-2) + 9
y = 2 + 9
y = 11

y = -(1) + 9
y = -1 + 9
y = 8

y = -(4) + 9
y = -4 + 9
y = 5

we got y every time so the answer is the first one

The answer is A) y = -x + 9

Hope this helps!

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3 years ago

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Given f(x)=5x+4, solve for x when f (x) = -6

Bingel [31]

So x=-2 after ndjsbdbbegshwoehev

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2 years ago

Ryan counted 63 Birds in his backyard last week this week he counted 71 Birds in his backyard about how many birds did Ryan coun

Airida [17]

Ryan counted 134 birds

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3 years ago

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Please help me out please

Bogdan [553]

A = 1/2aP

P = number of sides x length of each side

P = 8 x 5.8

P = 46.4

A= 1/2 x 7 x 46.4

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3 0

3 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