All categories

2 years ago

Second part need help ASAP

Mathematics

1 answer:

jarptica [38.1K]2 years ago

7 0

statements 13-L , 15-F , 6-G , and 10-C are true statements

hope this helps!

You might be interested in

Help please!!!!

Sonja [21]

Answer:

you already put this one

6 0

3 years ago

Solve the equation for the variable. <img src="https://tex.z-dn.net/?f=%20-%205v%20%2B%209%20%3D%20%20-%20v%20%2B%2017" i

AlekseyPX

Answer:

v= - 2

Step-by-step explanation:

...............

8 0

3 years ago

Sin A Cot A = Cos A, Cos A CosecA= Cot A, Sec A Cot A=CosecA, Cosec A Tan A =Sec A their r total 4 question. I will give brainli

EastWind [94]

Hey mate hope its help you.......

7 0

2 years ago

1. 232 is 8 times the sum of Mark's and his sister's ages. His sister is 21. Find Mark's age.

AURORKA [14]

Answer:

a. 8 years

b. $49.20

Step-by-step explanation:

The computation is shown below:

a. The mark age is

Given that

232 is 8 times of the sum of mark and his sister ages

And, his sister is 21

So,

Let us assume the mark age be x

now

8(x + 21) = 232

x + 21 = 29

x = 29 -21

x = 8

b. The original price of the socks is

As each pair get $3 off

And she bought 6 pairs for $31.20

= $31.20 + 6 × $3

= $31.20 + 18

= $49.20

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 : 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

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