Find the number of terms in this polynomial. 9p2 + 3p − 2

Peggy takes 2.5 hours to deliver 328 magazines. What is the rate per hour she is delivering the magazines

grandymaker [24]

Answer: 131 magazines per hour

Step-by-step explanation:

Peggy delivered 328 magazines in 2.5 hours.

To find out the rate at which she delivered these magazines, divide the number of magazines by the time taken to deliver them:

= Magazines / Time in hours

= 328 / 2.5

= 131 magazines per hour

3 0

3 years ago

PLEASE HELP

egoroff_w [7]

The temperature got colder by 11 regress

4 0

3 years ago

The following table shows the number of fire hydrants in each neighborhood overseen by District 9

choli [55]

Answer:

where is the table

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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 : p = 0.6

H₁ : p = 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 pI 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 pI 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

Write Two Different Addition Equations That Have 12 As The Solution.

kow [346]

6+6=12
8+4=12

or if you mean algebraic equation:

6+x=12
x+4=12

5 0

3 years ago

Read 2 more answers