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2 years ago

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Which rule describes the composition of transformations that maps pre-image ABCD to final image A"B"C"D"? Reflection across the

x-axis composition translation of negative 6 units x, 1 unit y. Translation of negative 6 units x, 1 unit y composition reflection across the x-axis. 90 degree rotation about point 0 composition translation of negative 6 units x, 1 unit y. Translation of negative 6 units x, 1 unit y composition 90 degree rotation about point 0.

1 answer:

NeTakaya2 years ago

7 0

Answer:

Your answer is the first point

A reflection across the x-axis and then a transformation of -6 and 1.

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preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. It was found tha

VladimirAG [237]

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

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

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

And rounded up we have that n=369

8 0

1 year ago

For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

natulia [17]

<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let $e$ the cost of the exam hours

Let $w$ be the cost of the workshop hours

Let $l$ be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

$e=2w$ equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

$w=2l$ equation (2)

Finally, we also know that 3hr exams 24hr workshops and 12hr lectures cost $528, so:

$3e+24w+12l=528$ equation (1)

Now, lets find the value of $l$ :

Step 1. Solve for $l$ in equation (3)

$3e+24w+12l=528$

$12l=528-3e-24w$ equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

$12l=528-3e-24w$

$12l=528-3(2w)-24w$

$12l=528-6w-24w$

$12l=528-30w$ equation (5)

Step 3. Replace equation (2) in equation (5) and solve for $l$

$12l=528-30w$

$12l=528-30(2l)$

$12l=528-60l$

$72l=528$

$l=\frac{528}{72}$

$l=\frac{22}{3}$

$l=7.33$

Cost of lectures = $7.33 per hour

3 0

1 year ago

Read 2 more answers

The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati

Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0

1 year ago

Alison is saving for retirement. Her company matches what she puts into her 401K in a ratio of 2:3. If she puts in $400 each mon

inessss [21]

Id say B...............

6 0

1 year ago

Read 2 more answers

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

Naily [24]

Answer:

Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in

Step-by-step explanation:

We are given;

n = 15

t-value = 1.66

Significance level;α = 0.05

So, DF = n - 1 = 15 - 1 = 14

From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14

Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.

8 0

1 year ago