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Schach [20]
2 years ago
10

The coordinates of the vertices of △PQR are P(1, 4) , Q(2, 2) , and R(−2, 1) . The coordinates of the vertices of △P′Q′R′ are P′

(−1, 4) , Q′(−2, 2) , and R′(2, 1) .
Which statement correctly describes the relationship between △PQR and △P′Q′R′ ?

△PQR is congruent to △P′Q′R′ because you can map △PQR to △P′Q′R′ using a translation 2 units to the left, which is a rigid motion.
△PQR is congruent to △P′Q′R′ because you can map △PQR to △P′Q′R′ using a reflection across the y-axis, which is a rigid motion.
△PQR is not congruent to △P′Q′R′ because there is no sequence of rigid motions that maps △PQR to △P′Q′R′ .
△PQR is congruent to △P′Q′R′ because you can map △PQR to △P′Q′R′ using a reflection across the x-axis, which is a rigid motion.

Mathematics
2 answers:
victus00 [196]2 years ago
7 0

Answer:

B

Step-by-step explanation:

sveticcg [70]2 years ago
5 0

ANSWER


The correct  answer is B


<u>EXPLANATION</u>

If we analyse the coordinates carefully, you realize there is a mapping of

(x,y)\rightarrow(-x,y)

This is a reflection in the y-axis.


Since the transformation is a reflection, the shape is preserved. Therefore, the image triangle P'Q'R' of triangle

PQR are congruent.



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The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr
9966 [12]

Answer:

(a) Perimeter = 32.2\ cm

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

Area = 64.8

Required

Calculate the perimeter.

Area is calculated as:

Area = L * W

Where

L = Length and W = Width

Substitute 64.8 for Area

64.8 = L * W

Make L the subject:

L = \frac{64.8}{W}

Perimeter is calculated as:

P = 2 * (L + W)

Substitute 64.8/W for L

P = 2 * (\frac{64.8}{W} + W)

P = \frac{129.6}{W} + 2W

To solve further, we take the derivative of P and set it to 0, afterwards.

dP = -\frac{129.6}{W^2} + 2

Set to 0

0 = -\frac{129.6}{W^2} + 2

Collect Like Terms

\frac{129.6}{W^2} = 2

Cross Multiply:

2W^2= 129.6

Divide through by 2

W^2 = 64.8

Take square roots

W = \sqrt{64.8

W = 8.05

Recall that:

L = \frac{64.8}{W}

So:

L= \frac{64.8}{8.05}\\

L= 8.05

The perimeter is:

Perimeter = 2 * (8.05 + 8.05)

Perimeter = 2 * (16.10)

Perimeter = 32.2\ cm

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

Tenths = \frac{1}{10}

So, the expression becomes:

((1 Group of 10 * \frac{1}{10}) and (1 group of 8 * \frac{1}{10}))/(6 groups of 3 * \frac{1}{10})

This gives:

((1 Group of \frac{10}{10}) and (1 group of \frac{8}{10}))/(6 groups of \frac{3}{10})

Group means product, so the expression becomes:

\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}

And, as used here means addition

\frac{(1 * \frac{10}{10} +  1 * \frac{8}{10})}{6 * \frac{3}{10}}

Simplify:

\frac{(1 * 1 +  1 * 0.80)}{6 *0.30}

\frac{(1 +  0.80)}{1.80}

\frac{1.80}{1.80}

= 1.00

7 0
2 years ago
Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 28% below the target pressure.
Alona [7]

Answer:

Step-by-step explanation:

Hello!

The monitoring system warn the driver when the tire pressure of the vehicle is 28% below target pressure.

Be X: target tire pressure of a certain car (pounds per square inch)

a)

X= 28 psi

If the monitoring system will warn the driver when the pressure is 28% below the target pressure: X-0.28X

First step, you have to calculate the 28% of 28psi

28*0.28= 7.84

Second step, is to subtract the calculated 28% to the target pressure:

28 - 7.84= 20.16

The TPMS will trigger a warning at 20.16 psi.

b)

If X~N(μ;σ²)

μ= 28psi (since the average is on target, then the target pressure for the car will be the average value of the distribution)

σ= 3psi

P(X≤20.16)

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now you have to look for the corresponding value of probability using the Z-table. Since the value is negative you have to the use the left entry of the Z-table, in the first column you'll find the integer and first decimal of the value -2.6- and in the first row you'll find the second decimal value -.-1

The value of probability that corresponds to -2.61 is:

P(Z≤-2.61)= 0.005

c)

You have to calculate the probability of inspecting a tire at random and it being inflated within recommended range, symbolically this is:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Calculate both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

The probability of the tire being inflated within recommended inflation range is 0.498.

I hope this helps!

5 0
2 years ago
The Ignorance Survey: United Kingdom A survey was conducted in the United Kingdom, where respondents were asked if they had a un
lukranit [14]

Answer:

Here we have given two catogaries as degree holder and non degree holder.

So here we have to test the hypothesis that,

H0 : p1 = p2 Vs H1 : p1 not= p2

where p1 is population proportion of degree holder.

p2 is population proportion of non degree holder.

Assume alpha = level of significance = 5% = 0.05

The test is two tailed.

Here test statistic follows standard normal distribution.

The test statistic is,

Z = (p1^ - p2^) / SE

where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]

p1^ = x1/n1

p2^ = x2/n2

p^ = (x1+x2) / (n1+n2)

This we can done in TI_83 calculator.

steps :

STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER

Test statistic Z = 1.60

P-value = 0.1090

P-value > alpha

Fail to reject H0 or accept H0 at 5% level of significance.

Conclusion : There is not sufficient evidence to say that  the percent of correct answers is significantly different between degree holders and non-degree holders.

7 0
2 years ago
On Thursday, Movietown Theater made 4/5 of their popcorn with butter. The remaining , 35.2 ounces was made without butter. How m
Y_Kistochka [10]
35.2 ounces is 1/5 of the popcorn

so 35.2 x 5 = 176

1 pound = 16 ounces 

176÷16=11

The movie theater made 11 pounds of popcorn
3 0
2 years ago
Read 2 more answers
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
2 years ago
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