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3 years ago

Solve the inequality −24z≥−13

Mathematics

2 answers:

kkurt [141]3 years ago

8 0

Answer:

z≤ 13 /24

all you have to do is divide both sides by 24

inysia [295]3 years ago

5 0

the actual answer is (-infinity sign, 13/24]

my computer wouldn't allow me to use the infinity sign. I'm on here trying to get answers for help and almost every last one has been wrong. people are paying for this site, so I wish we could stop putting unsure answers guys! Please and thanks!

Step-by-step explanation:

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Simplify the rational Expression <br>

qaws [65]

Answer:

$\frac{x^2+2x}{x-3}$

Step-by-step explanation:

We need to factor out numerator and denominator in order to simplify the rational expression by cancelling common factors.

Numerator : x^3 - 4 x = x (x^2 - 4) = x (x - 2) (x + 2)

Denominator (factoring by grouping):

x^2 - 5 x + 6 = x^2 - 3 x - 2 x + 6 = x (x - 3) - 2 (x - 3) = (x - 3) (x - 2)

Then we can cancel out the common factor (x - 2) in both numerator and denominator, leading to:

x (x + 2) / (x - 3) = (x^2 + 2)/ (x-3)

$\frac{x^2+2x}{x-3}$

4 0

3 years ago

Which equation is graphed in the diagram below

Deffense [45]

Answer:

y=2x+2

Step-by-step explanation:

7 0

3 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

4 years ago

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

7 0

3 years ago

Line L is perpendicular to a line with slope -5 and passes through (2, -2). Find an equation of line L and write the answer in t

irina [24]

Answer:

y=1x +8

Step-by-step explanation:

y-y1 = m(x-x1)

y+2 = -5(x-2)

y+2 = -5x + 10

y = -5x +8

Now to find line L we have to change the slope because it is perpendicular

y= 1x +8 (The slope is one fifth, but idk

5. to put it, I'm sorry)

4 0

3 years ago