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2 months ago

If x is a normal random variable with a standard deviation of 10, then 3x has a standard deviation equal to:

1 answer:

4 0

Standard deviation(σ $_{x}$ ) = √V(X) = 10
Therefore, the variance of X, V(X) = 10² = 100

The variance of aX is determined by
V(aX) = a²V(X)

Thus, the variance of 3X becomes
V(3X) = 3²V(X) = 9(100) = 900

Consequently, the standard deviation of 3X is

σ $_{3x}$ = √V(3X) = √900 = 30

Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

$H_0:p=0.08\\\\ H_a: p\neq0.08$

∵ $H_a$ this is a two-tailed test.

Test statistic:

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

, where p= denotes the population proportion

${\hat{p}$ = signifies the sample proportion

n= represents the sample size

Setting n= 6 and ${\hat{p}=\dfrac{6}{88}\approx0.068$ and p=0.08

$z=\dfrac{0.068-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{88}}}$

$z=\dfrac{-0.012}{0.0289199522192}\approx-0.415$

P-value for the two-tailed test:[2P(Z>|z|)

=2P(Z>|-0.415|)

=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]

=2(1-0.6609) [from the z-table]

=0.6782

Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.

This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.

8 0

2 months ago

To find the number in a square multiply the numbers in the two circles connected to it brainly

Leona [12618]

Response:

the solution can be found in the image

6 0

2 months ago

Describe the sample space for each of these experiments (a) A coin is tossed and a single die is rolled. (b) A paper cup is toss

PIT_PIT [12445]

Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space consists of all potential outcomes for each experiment:

a) The sample encompasses the outcomes resulting from tossing a coin (heads or tails) and throwing a die (a number between 1 and 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can settle in one of three different positions: on its top, bottom, or side:

S={top,bottom,side}

c) The entire set of cards within a standard deck constitutes the sample space:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

3 0

2 months ago

There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel are added to the mound. Two orders

PIT_PIT [12445]

Question is Incomplete; Complete question is given below;

A pile of g pounds of gravel exists in a quarry. During the day, 400 pounds of gravel is added to the pile. Two orders of 900 pounds are removed, resulting in the mound holding 1,500 pounds of gravel by day's end. Formulate the equation that accurately illustrates this scenario.

Answer:

The equation that best represents the situation is $g+400-1800=1300$ .