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2 years ago

3x sqaured-2x+3 (x=-4)

Mathematics

1 answer:

Effectus [21]2 years ago

6 0

Answer:

<h2> The answer is X= -4 </h2>

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Find the sum. Enter your answer in the box.<br><br> 3 + (–4)

Bess [88]

Answer:

the answer is -1

Step-by-step explanation:

3 0

2 years ago

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Use the cubic model y = 5a3 - 2a2 + a - 45 to find the value of y when x = 4.

Naddik [55]

Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.

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2 years ago

In ABC triangle AC=8, AB=5 and BC=6, AK=1. Find BK

In-s [12.5K]

Answer:

Bk=100

I' hope it's helpful for you

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1 year ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

2 years ago

Help !!!!!!!!!!!!!!!!

geniusboy [140]

Answer:

Hey there!

5(v+3)-10v>-10

5v+15-10v>-10

-5v+15>-10

-5v>-25

v<5

Let me know if this helps :)

5 0

3 years ago

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