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2 years ago

11

Given that a=4, b=3 and c=-4 work out b(squared) -2ac

2 answers:

Lesechka [4]2 years ago

4 0

Answer:

41

Step-by-step explanation:

b^2 -2ac

Let a=4 b=3 c=-4

( 3)^2 -2 (4) (-4)

Exponents first

9 - 2 (4) (-4)

Multiply

9 + 32

41

Monica [59]2 years ago

4 0

Answer:

41

Step-by-step explanation:

b^2 - 2ac plug in the values

3^2 -2(4)(-4) start with the exponent (remember PEMDAS)

9 -2(4)(-4) start multiplying from left to right

9 -8(-4) multiply

9 + 32 add

41 that's your answer :)

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If a data value in a normal distribution has a negative z-score, which of the following must be true?

mel-nik [20]

Correct Answer:
The data value must be less than the mean.

The formula for the z value is:

z-value = (Data value - Mean) / Standard Deviation

The z-value will be negative when the Numerator of the above expression will be negative. The numerator will be negative only when the Data Value is less than the mean value. For example, for a data value of 5 and mean value 8, the numerator will be -3, and when it will be divided by standard deviation, the resulting z value will be negative.

So, Third option is the correct answer.

4 0

3 years ago

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Solve the equation.<br> у+ 6 = -3y + 26<br> оооо<br> o y= -8<br> o y= -5<br> е у= 5<br> o y= 8

slega [8]

Y = 5

5+6
11

-3(5) +26
-15+26
11

11=11

5 0

3 years ago

Below is the graph of y=x^3. <br> Translate it to make it the graph of y=(x-4)^3-1

Vadim26 [7]

Answer:7=0.5 |x-4| -3

+3 +3

10=0.5 |x-4|

10/0.5=0.5/0.5 |x-4|

20=|x-4|

^

-20=x-4 20=x-4

+4 +4 +4 +4

-16=x 24=x

The answers are -16=x and 24=x

Hope this helps, an if it pleases you put as the brainliest answer

8 0

2 years ago

Read 2 more answers

9/25 ddivided by 729/1000

Nat2105 [25]

Kindly refer to attachment for solution.

Hope it helps ^_^

5 0

3 years ago

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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