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Leno4ka [110]
3 years ago
12

PLEASE HELP!!! this test will bring my grade up!! (question in picture)

Mathematics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

b

Step-by-step explanation:

I think it's SAS

It is because O is the common angle and the two sides are similar.

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The wall is to be 5ft by 7 ft. A brick is 4 inches by 12 inches. How many bricks are needed
Tatiana [17]

you need 83 bricks i believe

3 0
3 years ago
Miss Lim bought some puzzles for her pupils in the Maths Club. She bought a total
xxMikexx [17]

Answer:

She bought 24 Tangram puzzles

Step-by-step explanation:

Let the number of Tangram puzzles be t and the number of IQ puzzles be i

The sum of all the puzzles is 40;

i + t = 40 ••••••(i)

Cost of all Tangram puzzles at a cost of $4 per one will be;

4 * t = $4t

cost of all Iq puzzles at a cost of $2 per one is

2 * i = $2i

Sum

paid for all is $128

so;

4t + 2i = 128 •••••••(ii)

From equation i;

i = 40-t

Substitute into ii

4t + 2(40-t) = 128

4t + 80 - 2t = 128

2t = 128-80

2t = 48

t = 48/2

t = 24

6 0
2 years ago
Find the sum or difference.
KIM [24]

Answer:

the sum of the given matrices is

-7 -2

4 -15

6 0
2 years ago
An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish
otez555 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!

8 0
3 years ago
Blue Cab operates 15% of the taxis in a certain city, and Green Cab operates the other 85%. After a nighttime hit-and-run accide
NikAS [45]

Answer: The required probability is 0.414.

Step-by-step explanation:

Since we have given that

Probability of taxis in a certain city by Blue Cab P(B)= 15%

Probability of taxis by Green Cab P(G) = 85%

Let A be the event that eyewitness said that vehicle was blue.

P(A|B)=0.80

P(A|G)=0.80

P(A'|B)=0.20=P(A'|G)

Using the "Bayes theorem":

Probability that the taxi at fault was blue is given by

=\dfrac{P(B).P(A|B)}{P(B).P(A|B)+P(G).P(A'|G)}\\\\=\dfrac{0.15\times 0.8}{0.15\times 0.8+0.85\times 0.2}\\\\=\dfrac{0.12}{0.12+0.17}\\\\=\dfrac{0.12}{0.29}\\\\=0.414

Hence, the required probability is 0.414.

8 0
2 years ago
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