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3 years ago

Please I need help!!

Mathematics

1 answer:

Gelneren [198K]3 years ago

3 0

1 plus to the negative Peter plus the relative integers

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What is the difference of the ones and the tenth digit of the pi?

antiseptic1488 [7]

Answer:

Step-by-step explanation:

The value of pi is 22/7

Expressing as a decimal, the first ten digits of pi is 3.1415926535

The value in the tenth digit of the pi is 5 (tenth value after the decimal point)

The value of the ones is 3(The value before the decimal point)

The difference between both values = 5 - 3

The difference between both values = 2

Hence the difference of the ones and the tenth digit of the pi is 2

8 0

2 years ago

A dollar bill is worth _____ pennies. When written in decimal format, each penny can be written as ______ of a dollar.

Mkey [24]

One dollar is 100 pennies. In decimal for, each penny can be written as .01
.99 would be 99 cents. One more would make 1.00, or 1 dollar
Your answers are, (100, Hundredth)

4 0

2 years ago

Read 2 more answers

Someone please helppp ! i’ll give brainliest to the right answer !

Nuetrik [128]

Answer:

The function is 2 I believe

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Write the point-slope form of the given line that passes through the points (0, 3) and (4, 0). Identify (x1, y1) as the x-interc

Oksi-84 [34.3K]

The point slope form of the line has the following form:

y – y1 = m (x – x1)

The slope m can be calculated using the formula:

m = (y2 – y1) / (x2 – x1)

m = (3 - 0) / (0 – 4) = - ¾

So the whole equation is:

y – 0 = - ¾ (x – 4)

y = - ¾ (x – 4) or

7 0

3 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

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.

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

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-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

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago