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3 years ago

There are two packs of hyenas that live just outside of a small town in Harar, Ethiopia. Each night, the

Mathematics

1 answer:

notka56 [123]3 years ago

7 0

Answer:

yes

Step-by-step explanation:

the packs are fighting for a common goal against each other

so they are competing

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Solve for x I need help fast

Gre4nikov [31]

Answer:

4x-1=1/2(10x-20)

8x-2=10x-20

-2x=-18

X=9

3 0

3 years ago

at noon the temperature was 12 degrees it rose 8 during the afternoon and dropped 28 by midnight what was the temperature at mid

Reika [66]

thats impossible it cant drop 28 because 12+8 is 20 so since it rose 8 during the afternoon and dropped 20 by midnight that is inpossible

3 0

4 years ago

Read 2 more answers

Write the slope-intercept form of the equation of the line through the point (1,-3) with a slope of 2.

Mnenie [13.5K]

Answer:

The answer is

$\huge y = 2x - 5$

Step-by-step explanation:

To find an equation of a line when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have

$y + 3 = 2(x - 1) \\ y + 3 = 2x - 2 \\ y = 2x - 2 - 3$

We have the final answer as

$y = 2x - 5$

Hope this helps you

3 0

3 years ago

A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that wi

gulaghasi [49]

Answer:

a) 0.3585

b) 0.6415

c) 0.07548

Step-by-step explanation:

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 20

x = Number of successes required

p = probability of window frame with minor defect = 0.05

q = probability of window frame with no defect = 1 - 0.05 = 0.95

a) Probability that none of the frames picked will have defects

x = 0

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585

b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415

c) Probability that More than two will need adjustment = P(X > 2)

P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.

P(X > 2) = 1 - 0.9245 = 0.07548

7 0

4 years ago

A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta

Akimi4 [234]

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of women who find success with this new treatment = $\frac{7}{14}$ = 0.50

n = sample of women = 14

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the true proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ , $0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ ]

= [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

4 0

4 years ago