4. nine hundred forty-one thousand, eight hundred five

Combine like terms - 7 -3(4-2x)-10x

Shtirlitz [24]

Answer:

The simplified form would ve -19 - 4x

Step-by-step explanation:

Before we combine anything, we first must expand the expression by distributing the -3 to the terms in the parenthesis.

-7 - 3(4 - 2x) - 10x

-7 - 12 + 6x - 10x

Now we can combine the terms with and without variables.

-19 - 4x

8 0

3 years ago

Read 2 more answers

If you get a sneak peak at the salary list of your job and realize you are in the lowest bracket, which is the lowest 1%, what i

hoa [83]

Answer:

$z=-2.33$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary

Step-by-step explanation:

Let X the random variable that represent the salary, and for this case we can assume that the distribution for X is given by:

$X \sim N(48000,5500)$

Where $\mu=48000$ and $\sigma=5500$

And we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.99$ (a)

$P(X$ (b)

We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-2.33$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary

7 0

3 years ago

Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are

Musya8 [376]

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

7 0

3 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

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

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

2/3x = y -6 linear or nonlinear

djyliett [7]

Answer:linear

Step-by-step explanation:

If you go onto desmos and write the equation you get a straight line and it’s not curved

3 0

2 years ago

4. nine hundred forty-one thousand, eight hundred five​

4. nine hundred forty-one thousand, eight hundred five