The polygons are similar. Find the value of x. S X 1 9 the R X =

Gretta is 1 1/2 meters tall. Which of the following is equivalent to 1 1/2 meters?

yan [13]

Answer:

100 million

Step-by-step explanation:

7 0

4 years ago

Given the functions f(x) = 3(0.56)^x and g(x) = 6(4)^x

rodikova [14]

A) f(x) is decreasing because the base is less than 1.
0.56 is close to 0.5, so its like saying that you are taking half each time, therefore the value is getting smaller.

g(x) is increasing because the base is greater than 1.
you are multiplying by 4 each time, making the value bigger.

B ) The y-intercept is where x=0.
Anything to the '0' power is 1. Therefore the y-intercept is equal to the coefficient in front of each function.
f(x) = 3 , g(x) = 6

C) Just plug in x=4 to each function in a calculator.
f(4) = 0.295
g(4) = 1536

7 0

4 years ago

Find the solution(s) to 22 +57-3 = 0. Check all that apply.

irakobra [83]

Answer:

x = -3 and x = 1/2

Step-by-step explanation:

To find the solutions to

2x² + 5x - 3 = 0

we can use the quadratic formula, as follows:

$x = \frac{-b \pm \sqrt{b^2- 4(a)(c)}}{2(a)}$

$x = \frac{-5 \pm \sqrt{5^2- 4(2)(-3)}}{2(2)}$

$x = \frac{-5 \pm 7}{4}$

$x_1 = \frac{-5 + 7}{4}$

$x_1 = 0.5$

$x_2 = \frac{-5 - 7}{4}$

$x_2 = -3$

8 0

3 years ago

In five consecutive tosses of a coin, what is the probability of not getting a tail until the 5th coin toss?

jasenka [17]

The proabability would be 50% always

3 0

4 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

$P(X=280)=P(279.5$

8 0

3 years ago