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3 years ago

I’m so confused on shapes of pls help me

Mathematics

1 answer:

il63 [147K]3 years ago

8 0

Answer:

150 units

Step-by-step explanation:

2(2x)+2(x+12)=4x+2x+(x+3)

4x+2x+24=4x+2x+x+3

24-3=4x+2x+x-4x-2x

21=x

2(21)+(21)+12+2(21)+(21)+12

42+21+12+42+21+12

150

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An education firm measures the high school dropout rate as the percentage of 16 to 24 year olds who are not enrolled in school a

posledela

Answer:

The value of the test statistic $z = -0.6606$

Step-by-step explanation:

From the question we are told that

The high dropout rate is $\mu = 6.1$ % $= 0.061$

The sample size is $n = 1000$

The number of dropouts $x = 56$

The probability of having a dropout in 1000 people $\= x = \frac{56}{1000} = 0.056$

Now setting up Test Hypothesis

Null $H_o : p = 0.061$

Alternative $H_a : p < 0.061$

The Test statistics is mathematically represented as

$z = \frac{\= x - p}{\sqrt{\frac{p(1 -p)}{n} } }$

substituting values

$z = \frac{0.056 - 0.061}{\sqrt{\frac{0.061(1 -0.061)}{1000} } }$

$z = -0.6606$

6 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

Brainliest gets 22 points

Oliga [24]

Answer:

please provide a chart next time a question is asked

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Can someone help plz im giving brain

kenny6666 [7]

Answer:

for 20 the answer is B

Step-by-step explanation:

8 0

3 years ago

Question:

Artist 52 [7]

Ughhh I don’t know. I’ll try and solve it

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3 years ago