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3 years ago

I need help on the ones I did not do

Mathematics

1 answer:

Brut [27]3 years ago

7 0

Answer:

there is no documenty upload diffrent one

Step-by-step explanation:

You might be interested in

What is the solution to this equation? x + (- 14) = 4. (A) x = 10. (B) x = - 10. (C) x = - 18. (D) x = 18

Vika [28.1K]

First, you want to get x by itself on one side. We write the equation as $x-14=4$ . Since 14 is being subtracted from $x$ , we add it to both sides to cancel it out, $x-14+14=4+14$ or $x=18$ .

Your answer would be D.

4 0

3 years ago

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data

LuckyWell [14K]

Answer:

(25.53, 37.87): 95% CI

(23.59, 39.81): 99% CI

Step-by-step explanation:

The margin of error of a confidence interval is given by:

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

z is related to the confidence level. The higher the confidence level, the higher the values of z, and thus, we wider the confidence interval is.

In this question:

The narrower C.I. is the 95%, and the wider is the 99%. So

(25.53, 37.87): 95% CI

(23.59, 39.81): 99% CI

4 0

3 years ago

A pattern follows the rule "Starting with one, every consecutive line has a number one more than twice the previous line".

gtnhenbr [62]

Answer:

The fourth line will have 15 marbles

Step-by-step explanation:

The third line has 7 marbles. So, you double that and add one in order to find the number of marbles in the fourth line.

7*2=14

14+1=15

4 0

3 years ago

1/5x8 in simplest form

guapka [62]

Answer:

try the calculator best thing to use

4 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