Ask question

Ask question

All categories

andrey2020 [161]

3 years ago

15

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the rectangles formed by the sets of points

to their corresponding areas.
A(-9,8), B(-5,5), C(1,13), D(-3,16)

E(30,20), F(39,29), G(49,19), H(40,10)

I(-6,2), J(2,2), K(2,-8), L(-6,-8)

M(5,5), N(11,5), O(11,-5), P(5,-5)

Q(10,0), R(15,5), S(25,-5), T(20,-10)

U(0,5), V(15,20), W(25,10),X(10,-5)

50 square units

300 square units

100 square units

80 square units

2 answers:

tigry1 [53]3 years ago

6 0

IJKL=80 square units

QRST =100 square units

UVWX =300 square units

ABCD =50 square units

you are welcome!!!

lana [24]3 years ago

3 0

Answer:

Area of a rectangle = length × width

= Product of two adjacent sides

By the distance formula,

The length of a line segment having end points $(x_1,y_1)$ and $(x_2,y_2)$ ,

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\text{ unit}$

The area of rectangle having vertices A(-9,8), B(-5,5), C(1,13), D(-3,16),

= AB × BC

= √25 × √100

= 5 × 10

= 50 square unit,

The area of rectangle having vertices E(30,20), F(39,29), G(49,19), H(40,10)

= EF × FG

= √162 × √200

= 81√2 × 10√2

= 1620 square unit,

The area of rectangle having vertices I(-6,2), J(2,2), K(2,-8), L(-6,-8)

= IJ × JK

= √64 × √100

= 8 × 10

= 80 square unit,

The area of rectangle having vertices M(5,5), N(11,5), O(11,-5), P(5,-5)

= MN × NO

= √36 × √100

= 6 × 10

= 60 square unit,

The area of rectangle having vertices U(0,5), V(15,20), W(25,10),X(10,-5)

= UV × VW

= √450 × √200

= 300 square unit,

Area of rectangles having vertices Q(10,0), R(15,5), S(25,-5), T(20,-10)

= QR × RS

= √50 × √200

= 5√2 ×10√2

= 100 square unit.

You might be interested in

3/4 divided by 2 and 1/2??? Plssss thank youuu.

natita [175]

Answer:

3/4 divided by 2 1/2 is 0.3

7 0

2 years ago

Let’s assume that human body temperatures are normally distributed with a mean of 98.20° F and a standard deviation of 0.62° F.

uysha [10]

Answer: It is such a small percentage of the population that would be considered to have fever at that level.

So, it is not appropriate stringent standard.

Step-by-step explanation:

Since we have given that

Mean = 98.20°F

Standard deviation = 0.62°F

If Bellevue Hospital in New York City uses 100.6°F as the lowest temperature considered to be a fever.

So, $\bar{X}=100.6^\circ F$

So, using the normal distribution, we first find the value of z.

$z=\dfrac{\bar{x}-\mu}{\sigma}\\\\x=\dfrac{100.6-98.2}{0.62}\\\\z=3.87$

Since z = 3.87

So, p = 0.0001 =0.1%

So,It is such a small percentage of the population that would be considered to have fever at that level.

So, it is not appropriate stringent standard.

4 0

3 years ago

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam

soldi70 [24.7K]

Answer:

a) $52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) $52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

c) $52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) $52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

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".

$\bar X =52.8$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

$52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

So on this case the 95% confidence interval would be given by (51.554;54.046)

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that $t_{\alpha/2}=2.71$

Now we have everything in order to replace into formula (1):

$52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

$52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

So on this case the 99% confidence interval would be given by (51.129;54.471)

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

3 0

3 years ago

A machine packs at a constant rate of 2/3 of a box every 1/2 minuet. What’s is the number of boxes per minute that’s the machine

lions [1.4K]

For this case what we can do is the following rule of three:
2/3 of a box ------> 1/2 minutes
x of a box ---------> 1 minute
Clearing x we have:
x = (1 / (1/2)) * (2/3)
Rewriting:
x = 2 * (2/3)
x = 4/3
Answer:
the number of boxes per minute that the machine packs is:
x = 4/3

4 0

3 years ago

You and your friend are playing a game. The two of you will continue to toss a coin until the sequence HH or TH shows up. If HH

miskamm [114]

Answer:

The probability of friend A winning with HH = 1/4.

Step-by-step explanation:

The probability of an event, A is P(A) given by the relationship;

P(A) = (The number of required outcome)/(The number of possible outcomes)

The parameters given are;

The condition of friend A winning = Coin toss sequence HH shows up

The condition of friend B winning = Coin toss sequence TH shows up

The number of possible outcomes = TT, TH, HH, HT = 4

(TH and HT are taken as different for the game to be fair)

The number of required outcome = HH = 1

Therefore;

The probability of friend A winning with HH = 1/4.

5 0

2 years ago