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2 years ago

7

What does 8.3 × 10^ -5 in standard notation

1 answer:

xz_007 [3.2K]2 years ago

6 0

Answer:

0.000083

Step-by-step explanation:

When you multiply by 10^-5, you're really multiplying by 0.00001. This multiplied by 8.3 is 0.000083. It also means that you just move the decimal place in 8.3 to the left five times, where you'll get the same answer.

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R= 4/3(p-q) <br><br>solve for p

Naya [18.7K]

Answer = p=12-4 -12=17=18+19

p-q

p=12

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2 years ago

I need help with this. I'm stuck. PLEASE SAVE ME

vagabundo [1.1K]

Answer: Its the one.

Step-by-step explanation:

Your going to subtract 2 from each sides making the equation, -5 is less than or equal to 25, then your gonna divide -5 by 25, which will give you -5, and then the sign will switch to greater than or equal to because you divided by a negative number. The dot on the number line will be solid because the sign is greater than or equal to. You're welcome :)

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2 years ago

Solve 7p - 4 + 12p = -3 (5 + p) Can you show the work please?

prisoha [69]

7p - 4 + 12p = -3(5 + p)
7p - 4 + 12p = -15 - 3p
+ 4 + 4
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+ 3p + 3p
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p = -1/2 Answer

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3 years ago

Read 2 more answers

An arc created by a central angle has the same measure as the central angle. True False

KATRIN_1 [288]

You're answer is True

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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