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3 years ago

11

Write an equivalent expression of 16x - 24

1 answer:

telo118 [61]3 years ago

8 0

Answer:

2x-3

Step-by-step explanation:

16/8=2

24/8=3

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1/2(8x-6) = x + 9 solve ... hurry

QveST [7]

Answer:

x=4

Step-by-step explanation:

4x-3=x+9

4x-x-3=9

4x-x=9+3

3x=9+3

3x=12

3/3

12/3=4

x=4

5 0

3 years ago

Read 2 more answers

What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?

Effectus [21]

$\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}$

$Answer:\ \dfrac{19}{52}$

5 0

4 years ago

Please help asap ill mark you brainliest!!

julia-pushkina [17]

Answer:

4,184.87 square meters

Step-by-step explanation:

let me know if you want to see work . . .

3 0

3 years ago

How to solve x9+y10=4

Taya2010 [7]

Answer:

if you meant 9x+10y=4 then the x intercepts are (4/9,0) and the y intercepts are (0,2/5).

Step-by-step explanation:

5 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