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3 years ago

Find the mean 78 77 76 75 74

Mathematics

1 answer:

serious [3.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

(78+77+76+75+74)/5=x

x=76.

So the mean is 76.

Hope this helps plz hit the crown :D

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Write the equation of the line that passes through the point (-6, 1) and has a slope of 1/2

Likurg_2 [28]

Answer:

y = x + 2

i think this is helpful

Step-by-step explanation:

equation= y - 1 = 1

x - -6 2

then you close multiplication

= 2(y -1)= 1(x - -6)

2y - 2 = x - -6

you make Y as a subject

2y = x- -6 + 2

2y = x + 4

then you divide by 2 on both sides

2y = x + 4

2 2 2

y = x + 2

8 0

4 years ago

Solve the equation by factoring. 5z2 – 21z + 4 = 0

wariber [46]

(5z-1)(z-4)
0= 4 and 1/5

3 0

4 years ago

Which choice shows the product of 22 and 39?

Yuliya22 [10]

Answer: (22)(39) = 858.

Step-by-step explanation:

Product of a number simply means that one has to multiply the numbers that are given. For example, if a student is told to find the product of 2 and 5, it simply means that 2 × 5 = 10.

Regarding the above question, the product of 22 and 39 is represented by: (22)(39) = 858. It should be noted that 22 × 39 = 858

8 0

3 years ago

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are list

Kitty [74]

Answer:

It does not appear that, as a group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1067}{15} = 71.13$

Sum of squares of differences = 4739.733

$S.D = \sqrt{\frac{4739.733}{14}} = 18.39$

Population mean, μ = 60 minutes

Sample mean, $\bar{x}$ = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}$

We use Two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34$

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear that, as a group, the students are reasonably good at estimating one minute.

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3 years ago

Martin is comparing two different video game stores.

kondor19780726 [428]

He would save $4.34 every game because when you always

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3 years ago