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morpeh [17]
3 years ago
12

(×-0.20×)+0.10 (×-0.20×)

Mathematics
2 answers:
Rufina [12.5K]3 years ago
5 0
0.88x distributive property and isolate the variables
Anna71 [15]3 years ago
3 0
Simply the expression

(x-0.2x)+0.10 (x-0.20x)

=(0.8x)+0.1+0.8x

=0.8x+0.1+0.8x

=(1+0.1) × 0.8x

=1.1 × 0.8x

=0.88x
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Calculate x Give your answer correct to 1 d.p. Х 35° 15 cm<br> it is a trigonometry question.​
OlgaM077 [116]

Here we are finding x, given the angle and adjacent side. To find x we will use the function cos as cos = a/h.

So let's do cos(35°) = 15cm / x

cos(35°) = 0.8 (1 dp)

x = 15 cm / cos(35°)

x = 18.3 cm (1 dp)

6 0
2 years ago
a carpenter has a wooden stick that is 84 centimeters long. She cuts off 25% from the end of the stick. then she cuts the remain
Karo-lina-s [1.5K]
She cuts off 21 cm. this leaves 63cm. cut into 6 equal pieces gives 10.5 cm pcs
3 0
2 years ago
If a cow has a mass of 9×10^2 kilograms, and a blue whale has a mass of 1.8×10^5 kilograms, which of these statements is true?
trasher [3.6K]

Answer:

The 1st answer, <em>The blue whale has about 200 times more mass. </em>might be correct.

Step-by-step explanation:

9 x 10 ^2= 900 kilograms

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180000 kilograms / 900 kilograms= 200 times

5 0
2 years ago
The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr
aniked [119]

Answer:

(a) E(X) =  0.383

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) P(x = 4) = 0.000169

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$ E(X) =  \frac{n \times r}{N} $

$ E(X) =  \frac{10 \times 8}{209} $

$ E(X) =  \frac{80}{209} $

E(X) =  0.383

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$ P(x = 4) =  \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}} $

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$ P(x = 4) =  \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{70 \times 84944276340}{35216131179263320}

P(x = 4) = 0.000169

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0
2 years ago
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Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

If two lines coincide then the system of linear equations has infinitely many solutions.

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2 years ago
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