HI HELP ITS A QUIZZZ!

Please show steps to this problem

Leto [7]

Answer:

Answer is D. 1572 square units.

4 0

3 years ago

Ayliah is 7 years more than 1/2 of Deb's age<br> use x for the variable

irina [24]

Answer:

The expression $\frac{1}{2}x+7$ can be used to find Ayliah's age.

Step-by-step explanation:

Given statement is:

Ayliah is 7 years more than 1/2 of Deb's age.

Let,

x represent Deb's age.

The expression for Ayliah's age would be;

7 more means addition.

Ayliah's age = $\frac{1}{2}x+7$

Hence,

The expression $\frac{1}{2}x+7$ can be used to find Ayliah's age.

7 0

2 years ago

HELP!

tigry1 [53]

Answer:

$b=40.01\ units$

Step-by-step explanation:

we know that

Applying the law of sines

$\frac{b}{sin(B)}=\frac{a}{sin(A)}$

$b=\frac{a}{sin(A)}(sin(B))$

we have

$a=25\ units$

$A=32.5\°$

Find the measure of angle B

Remember that the sum of the interior angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

substitute the given values

$32.5\°+B+26.8\°=180\°$

$B=180\°-59.3\°=120.7\°$

Find the length side b

$b=\frac{a}{sin(A)}(sin(B))$

substitute the values

$b=\frac{25}{sin(32.5\°)}(sin(120.7\°))$

$b=40.01\ units$

8 0

3 years ago

Solve for the variable m. <br>m - n = 5

Ber [7]

7=m

2=n

so it will be 7-2=5

7 0

2 years ago

Read 2 more answers

Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. a

Y_Kistochka [10]

Answer:

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

$np = 37 *0.2 = 7.4 \geq 5$

$n(1-p) = 37*0.8 = 29.6\geq 5$

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

7 0

2 years ago