The tension t in a spring varies directly with the distance d the spring is stretched. If the
1 answer:
Answer:
t = 21d
Step-by-step explanation:
Here tension and distance are denoted t by d and respectively. Since t varies directly with d , then the relation between t and d is k=td Given that, for the tension 42 pounds, the spring is stretched 2 inches. Substituting t = 42 and d= 2 into t = kd find the value of k . Therefore,
42= K (2)
K= 21
Thus the relationship is t = 21d
(Hope this helps can I pls have brainlist (crown)☺️)
You might be interested in
Answer:
The length of BC is needed because it is the side opposite ∠A.
Step-by-step explanation:
Given the right angles triangle as shown in the attachment, we can get sin(A) without using Pythagoras theorem. Instead we will use SOH CAH TOA trigonometry identity.
According to SOH:
Sin(A) = Opposite/Hypotenuse
Sin(A) = |BC|/|AB|
Opposite side of the triangle is the side facing ∠A.
Based on the formula, we will need to get the opposite side of the triangle which is length BC for us to be able to determine sinA since the hypotenuse is given.
Answer:
tons of high steel and
tons of low steel
Step-by-step explanation:
Let x be the number of tons of high grade steel and y be the number of tons ow low grade steel needed.
In x tons of high grade steel there are
tons of iron
tons of magnese
In y tons of low grade steel there are
tons of iron
tons of magnese
NASA orders 500 tons of steel, so
and specifies that it must be in the proportion 80% iron and 20% magnese, so
From the first equation,
Substitute it into the second equation:
Answer:
a) The expected value of the sample mean weight is 20.4 pounds.
b)The standard deviation of the sample mean weight is 0.123.
c) There is a 14.46% probability the sample mean weight will be less than 20.27.
d) This value is .
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that
Consider the sample mean weight of 100 watermelons of this variety. This means that .
a. What is the expected value of the sample mean weight? Give an exact answer.
By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.
b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.
By the Central Limit Theorem, that is:
The standard deviation of the sample mean weight is 0.123.
c. What is the approximate probability the sample mean weight will be less than 20.27?
This is the pvalue of Z when .
Since we are working with the sample mean, we use instead of
in the Z score formula
has a pvalue of 0.1446.
This means that there is a 14.46% probability the sample mean weight will be less than 20.27.
d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?
This is the value of that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.
So we use
This value is .