A-(<br> b.(h)126= (4x 1)(x 1)

I NEED HELP plz i will pay u 15 points if u help meeee

Arlecino [84]

Answer:

1. 9/10

2. 14/15

3. 1 1 /6 or 7/6

Step-by-step explanation:

1. Change the 5, the denominator to 10. 8/10

then add to 1/10 8/10 + 1/10= 9/10

2. Change both denimnators to 15. You will then get 9/15 and 5/15.

Add them together and get 14/15

3. Change both denominators to 6 and you will get 3/6 and 4/6.

Add them together to get 7/6.

If you want to make it into a mixed fraction it would be 1 1/6 (as in one and one-sixth)

8 0

2 years ago

Read 2 more answers

I have been having trouble doing this problem 5x + 34

Andru [333]

Answer:

Hey! Well you cant simplify it anymore.

Step-by-step explanation:

You cant add an unknown number to another number. So its still 5x+34

I Hope this helps :|

3 0

2 years ago

Find the value of the trigonometric ratio.<br> 9) sin z

azamat

Answer:

SinZ=24/40 = 12/20 = 3/5

C)

3 0

2 years ago

8- 5 (a + b) + 3 (a - 2b)

Alexeev081 [22]

Answer: 8-2a-11b

.....

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

A-( b.(h)126= (4x 1)(x 1)