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2 years ago

Please answer this friends I am stuck ?

Mathematics

1 answer:

goblinko [34]2 years ago

4 0

Answer:

10°C average temperature

Step-by-step explanation:

Average Temperature = Lowest + 17

= -7 + 17

= 17 - 7

= 10°C average temperature

You might be interested in

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

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 $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

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:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

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 $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

$Z = \frac{X - \mu}{s}$

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ 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 $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

$Z = \frac{X - \mu}{s}$

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago

Mia is making costumes for a play. Each costume needs 3 7/9 yards of fabric. If Mia is making 6 costumes, how much fabric does s

mihalych1998 [28]

If she is making 6 costumes that need 3 7/9 yards each, the equation would be 6 x 3 7/9.

The answer to that equation would be 22 2/3.

How to get the answer:
3 x 6 = 18
7 x 6 = 42 (the seven is from the 7/9)
42/9 = 4 2/3

then add the eight teen and the 4 2/3 and you get 22 2/3.

3 0

2 years ago

For pigs, the length of pregnancies varies according to a normal distribution with a mean of 114 days and a standard deviation o

riadik2000 [5.3K]

Answer: The length of approximately 68% of all pig pregnancies will fall between 109 days and 119 days.

Step-by-step explanation:

According to the empirical rule , 68% of the population falls within one standard deviations from the mean.

Given : For pigs, the length of pregnancies varies according to a normal distribution with a mean of 114 days and a standard deviation of 5 days.

According to the Empirical Rule, the length of approximately 68% of all pig pregnancies will fall between $114-1(5)$ days and $114+1(5)$ days .

i.e. the length of approximately 68% of all pig pregnancies will fall between 109 days and 119 days.

7 0

2 years ago

Find (2x+4) (x-4) using the table of products. write your answer in standard form.

ValentinkaMS [17]

Answer:

2x^2-4x-16

Step-by-step explanation:

6 0

2 years ago

What is the y-intercept of the line represented by the equation y = 2x - 3?

cestrela7 [59]

Y =Mx + C
where M = gradient / slope
C = y intercept

comparing with y = 2x - 3
M = 2
C = -3

5 0

2 years ago