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Anvisha [2.4K]
3 years ago
13

42 is the product of 6 and y. Solve for y

Mathematics
2 answers:
Molodets [167]3 years ago
7 0

Answer:

y=7

Step-by-step explanation:

42/6=7

fgiga [73]3 years ago
4 0

Answer:

y =7 iits divided

Step-by-step explanation:

y =6/42

y= 7

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On an August day in Fairbanks, Alaska, the temperature changed by an average of 1.5° C per hour from its low of 5° C at 5:00 am.
zhuklara [117]

Answer:

9.5 C?

Step-by-step explanation:

im guessing since it's last change was an additional 4.5

5 0
3 years ago
Read 2 more answers
Someone help me please
Masteriza [31]

Answer:

(5,1)

Step-by-step explanation:

Hi there!

Midpoint = \displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} ) where the two endpoints are (x_1,y_1) and (x_2,y_2)

Plug in the given information:

Midpoint = (5,3), Endpoint = (5,5)

(5,3)=\displaystyle (\frac{5+x_2}{2},\frac{5+y_2}{2} ) where  (x_2,y_2) is the other endpoint

Solve for x_2:

\displaystyle \frac{5+x_2}{2} =5\\\\5+x_2 =10\\x_2 =5

Solve for y_2:

\displaystyle\frac{5+y_2}{2}=3 \\\\{5+y_2}=6\\y_2=1

Therefore, the other endpoint (x_2,y_2) is (5,1).

I hope this helps!

4 0
3 years ago
Read 2 more answers
g A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are he
stira [4]

Answer:

Step-by-step explanation:

Corresponding heights of presidents and height of their main opponents form matched pairs.

The data for the test are the differences between the heights.

μd = the​ president's height minus their main​ opponent's height.

President's height. main opp diff

191. 166. 25

180. 179. 1

180. 168. 12

182. 183. - 1

197. 194. 3

180. 186. - 6

Sample mean, xd

= (25 + 1 + 12 - 1 + 3 + 6)/6 = 5.67

xd = 5.67

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (25 - 5.67)^2 + (1 - 5.67)^2 + (12 - 5.67)^2+ (- 1 - 5.67)^2 + (3 - 5.67)^2 + (- 6 - 5.67)^2 = 623.3334

Standard deviation = √(623.3334/6 sd = 10.19

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 6 - 1 = 5

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (5.67 - 0)/(10.19/√6)

t = 1.36

We would determine the probability value by using the t test calculator.

p = 0.12

Since alpha, 0.05 < than the p value, 0.12, then we would fail to reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that for the population of heights for presidents and their main​ opponents, the differences have a mean greater than 0 cm.

5 0
3 years ago
Kate sets a goal to walk 3 miles every day for a year. How many miles more or less than 1000 miles will she walk in the year
Law Incorporation [45]

Answer:

95

Step-by-step explanation:

365x3=1095


6 0
3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit theorem

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
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