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2 years ago

6

Help me with this question please!! :)

1 answer:

Anna11 [10]2 years ago

7 0

Answer: 9516793/9765625

Step-by-step explanation:

We can subtract the probability none of them have been vaccinated from 1.

If 52% of people are vaccinated, then 48% of people are not vaccinated.

Thus, the probability is $1-(0.48)^{5} =\frac{9516793}{9765625}$

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If roberto can build 1 house in 23 days, how many houses can he build in 207 days?

rusak2 [61]

He would be able to build 9 houses.

$\frac{207}{23} = 9$

3 0

3 years ago

The sum of the measures of two complementary angles exceeds the difference of their measure by 86_. Find the measure of each ang

Bad White [126]

As the sum of two complementary angles = a + b = 90 degree -------(1)

Difference of two complementary angles = a – b

According to question;

The difference of two complementary angles exceed the sume (90) by 86.

So

a – b = 90 – 86 = 4 -------(2)

adding equation (1) and (2);

2a = 90 + 4 = 94

a = 94 / 2 = 47

putting a = 47 in equation (1);

47 + b = 90

b = 90 – 47 = 43

<span>So angle a = 47 and angle b = 43</span>

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4 years ago

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

<em /> $S_{1893} =5632.98$ <em />

<em />

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

<em /> $S_{1893} =5632.98$ <em> ----- approximated</em>

4 0

3 years ago

What are the intercepts if the graphed functions

KonstantinChe [14]

The intercepts represent the points where the graph touches the axis the given. With that said, your y intercept is (0, -3) and the x intercept is (-1,0).

6 0

3 years ago

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find the smallest number by which 368 be multiplied so that the product is a perfect square find the square root of the new squa

Vikki [24]

Answer:

0

Step-by-step explanation:

368×0=0

Now,

√0=0

0 is the perfect square, it is smallest number and it is your answer.

8 0

3 years ago

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