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AleksandrR [38]

3 years ago

14

A medical device company knows that 11% of patients experience injection-site reactions with the current needle. If 4 people rec

eive injections with this type of needle, what is the probability that none of the 4 people get an injection-site reaction? 0.0001 0.3726 0.4400 0.6274

1 answer:

Orlov [11]3 years ago

6 0

Answer: 0.6274

Step-by-step explanation:

Given: The proportion of patients experience injection-site reactions with the current needle : p=0.11

Sample size : n= 4

Let x be a binomial random variable that represents the people get an injection-site reaction.

Binomial probability formula: $P(X=x)= ^nC_x p^x(1-p)^{n-x}$

The required probability : P(x=0)

$=\ ^4C_0(0.11)^0(1-0.11)^4\\\\=(1)(1)(0.89)^4\\\\=0.62742241\\\approx0.6274$

Hence, the probability that none of the 4 people get an injection-site reaction = 0.6274

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The average of eight numbers is 56. When one of the numbers was left out, the mean decreased to 54. What number was left out?

Alexandra [31]

The number that was left out is 70.

<h3>What is the number that was left out?</h3>

Average is the sum of a set of numbers divided by the total numbers in the data set.

Average = sum of numbers / total numbers

Sum of numbers when there are 8 numbers = 56 x 8 = 448
Sum of numbers when there are 7 numbers = 54 x 7 = 378
Difference = 448 - 378 = 70

To learn more about average, please check: brainly.com/question/25842202

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4 0

2 years ago

Solve the value for v.

NemiM [27]

Answer:

Hey mate !!

Your answer is so simple :)

<h2>V= 9°</h2>

Go through the explanation to understand.

Step-by-step explanation:

In the given question,

We see that 105° and it's adjacent angle (i.e. 8v+3° ) forms a supplementary angle. Which means the sum of 8v+3°+105° is 180°

<h3>Therefore,</h3>

According to the question

$105 + 8v + 3 = 180$

$= > 105 + 8v = 180 - 3$

$= > 8v = 177 - 105$

$= > 8v = 72$

$= > v = \frac{72}{8}$

$= > v = 9$

Answer the value of v is 9° .

Hope it helps you!!

#IndianMurga ツ

3 0

2 years ago

There are 80 people waiting to tour the science museum. The tour guide takes 4 groups of 16 people each. The remaining people wi

Anna11 [10]

Answer:

16 people

Step-by-step explanation:

Given that :

Number of people to tour = 80

Number per group = 16

Number of groups = 4

Total number of people who make up the entire groups :

Number per group * number of groups

16 * 4 = 64

The number of people in the last tour group :

80 - 64

= 16 people

7 0

3 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u> $p_1-p_2$ <u>)</u><u> is ;</u>

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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

<u>95% confidence interval for</u> ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

How do I Solve this problem?

nlexa [21]

U would factor by grouping and get x^2(x-2) + 3(x-2) and then get (x^2 +3) (x-2)

6 0

3 years ago

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