1) A group of 32 students and 8 adults are going on a field trip to the recycling center

The ratio of the lengths of the sides of △ABC is 3:6:7. M, N, and K are the midpoints of the sides. Perimeter of △MNK equals 7.4

artcher [175]

Answer:

AB=2.775

BC=5.55

CA=6.475

Step-by-step explanation:

Since midpoints split their sides in half, we can see that the triangle MNK formed by the midpoints will be half the perimeter of the triangle ABC. Since P of MNK = 7.4, we know that the perimeter of ABC = 7.4 * 2, which is 14.8. Now we can split the 14.8 so that it follows the ratio.

3+6+7=16

14.8/16=0.925

AB=0.925*3=2.775

BC=0.925*6=5.55

CA=0.925*7=6.475

8 0

2 years ago

Two pounds of sugar cost $1.40. How much sugar do you get per dollar? Round your answer to the nearest hundredth, if necessary.

Vsevolod [243]

Answer:

1.429 lbs per dollar of sugar

Step-by-step explanation:

multiply by 5 get 10 lbs of sugar for 7 dollars divide by 7 get 1.4285 round up

5 0

2 years ago

A quality-control inspector is testing a batch of printed circuit boards to see wheater they are capable of performing in a high

avanturin [10]

Answer:

see explaination

Step-by-step explanation:

Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.

a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.

b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.

6 0

2 years ago

The system shown has the unique solution (2, y, z). Solve the system and select the values that complete the solution. y = 0 y =

madreJ [45]

So we are given a system:
$3x-2y+3z=0\\
-3x - 5y - 5z= -21$
Substitute x = 2 we get the system:
$-2y+3z=-6\\
- 5y - 5z= -15$
Multiply the first equation by -5 and the second by 2 we get the system:
$10y-15z=30\\
- 10y - 10z= -30$
Adding the two equations we get :
$-25z=0\text{ then}z=0.$
We find the value of y by using any of the other equations like this:
$-2y=-6\\y=3.$
Final solution:
$z=0,y=3$

8 0

2 years ago

Read 2 more answers

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

2 years ago