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3 years ago

The ordered pair (22, -45) is located in which quadrant of a coordinate plane?

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

Answer:

where is the graph

Step-by-step explanation:

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Construct a new equation by adding 5x + 5y = 35 to 3.x - 2y = 1

ohaa [14]

5x + 5y + 3x - 2y = 36
8x + 3y = 36

6 0

3 years ago

Please Help!! Thanks in advance!

galben [10]

Given:

The equation is:

$\dfrac{y}{p-q}=\dfrac{y}{p+q}$

To find:

The value of y.

Solution:

We have,

$\dfrac{y}{p-q}=\dfrac{y}{p+q}$

On cross multiplication, we get

$y(p+q)=y(p-q)$

$yp+yq=yp-yq$

$yp+yq-yp+yq=0$

$2yq=0$

Dividing both sides by 2q, we get

$\dfrac{2yq}{2q}=\dfrac{0}{2q}$

$y=0$

Therefore, the correct option is A.

7 0

3 years ago

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio

Reptile [31]

Answer:

The probability is $P(|p-\^{p}| > 0.03) = 0.0040$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.07$

The mean of the sampling distribution is $\mu_p = 0.07$

The sample size is n = 600

Generally the standard deviation is mathematically represented as

$\sigma_p = \sqrt{\frac{p (1 -p)}{n} }$

=> $\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }$

=> $\sigma_p = 0.010416$

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

$P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )$

Now add p to both side of the inequality

=> $P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )$

Now converting the probabilities to their respective standardized score

=> $P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )$

=> $P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)]$

From the z-table

$P(Z \le 2.88) = 0.9980$

and

$P(Z \le -2.88) = 0.0020$

$P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020]$

=> $P(|p-\^{p}| > 0.03) = 0.0040$

7 0

3 years ago

A 1.5 kilogram lab cart accelerated unitorily from rest to a speed of 2.0 meters per second in 0.50 second What is the magnitude

NISA [10]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning.

The cart's acceleration is

(0 - 2 m/s) / (0.3 sec)

= ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .

Newton's second law of motion says

Force = (mass) x (acceleration) .

For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)

= ( - 1.5 x 20/3 ) (kg-m/s²)

= - 10 newtons .

The force is negative because it acts opposite to the direction
in which the cart is moving, it causes a negative acceleration,
and it eventually stops the cart.

3 0

3 years ago

Let f be a function of two variables that has continuous partial derivatives and consider the points

Sergeu [11.5K]

Answer:

The directional derivative of f at A in the direction of $\vec{u}$ AD is 7.