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Zigmanuir [339]

3 years ago

14

Anna is in a cave 39 feet below the cave entrance. She descends 19 feet, then ascends 25 feet. Find her new position relative to

the cave entrance.

1 answer:

Ne4ueva [31]3 years ago

6 0

-40-13=-53

-53+18=-35

35 feet below the cave entrance

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If y varies directly with x, and y = 2 when x = −4, what is the value of y when x = 20

Klio2033 [76]

Answer:

y = - 10

Step-by-step explanation:

Given that y varies directly with x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 2 when x = - 4, that is

2 = - 4k ( divide both sides by - 4 )

= k, that is

k = -

y = - x ← equation of variation

When x = 20, then

y = - × 20 = - 10

3 0

3 years ago

Avoiding an accident while driving can depend on reaction time. Suppose that reaction time, measured from the time the driver fi

Nadya [2.5K]

Answer:

2.5% of drivers have a reaction time more than 2.14 seconds

16% of drivers have a reaction time less than 1.78 seconds

84% of drivers have a reaction time less than 2.02 seconds

Step-by-step explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

The normal distribution is symmetric, which means that 50% of the measures are below the mean and 50% are above.

In this problem, we have that:

Mean = 1.9s

Standard deviation = 0.12

What percentage of drivers have a reaction time more than 2.14 seconds?

2.14 = 1.9 + 2*0.12

So 2.14 is two standard deviations above the mean.

Of the 50% of the measures above the mean, 95% are within 2 standard deviations of the mean, so, below 2.14. The other 5% is above.

0.05*0.5 = 0.025

2.5% of drivers have a reaction time more than 2.14 seconds

What percentage of drivers have a reaction time less than 1.78 seconds?

1.78 = 1.9 - 0.12

So 1.78 is one standard deviation below the mean.

Of the 50% of the measures that are below the mean, 68% are within one standard deviation of the mean, that is, greater than 1.78.

100 - 68 = 32

0.32*50 = 0.16

16% of drivers have a reaction time less than 1.78 seconds

What percentage of drivers have a reaction time less than 2.02 seconds?

2.02 = 1.9 + 0.12

So 2.02 is one standard deviation above the mean.

Of the measures that are below the mean, all are below 2.02.

Of those that are above, 68% are below 2.02.

0.5 + 0.68*0.5 = 0.84

84% of drivers have a reaction time less than 2.02 seconds

5 0

3 years ago

Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert

adell [148]

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

$Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}$ $= 60.67$

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: $s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}$

Put value in formula of Standard Deviation,

$s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}$ = 40.75

Step-3: Then, we have to find the critical value by chi-square.

$X_{1-\alpha/2}^{2}=3.82$

$X_{1-\alpha/2}^{2}=21.92$

Then, find the confidence interval which is 95%.

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9$

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9$

5 0

3 years ago

Using the binomial theorem , obtain the expansion of :

andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1 6 15 20 15 6 1

Decreasing powers of 1 from $1^{6}$ to $1^{0}$

Increasing powers of 3x from $(3x)^{0}$ to $(3x)^{6}$

$1+3x)^{6}$

= 1. $1^{6}$ $(3x)^{0}$ + 6. $1^{5}$ $(3x)^{1}$ + 15. $1^{4}$ $(3x)^{2}$ + 20. $1^{3}$ $(3x)^{3}$ + 15.1² $(3x)^{4}$ + 6. $1^{1}$ $(3x)^{5}$ + 1. $1^{0}$ $(3x)^{6}$

= 1 + 18x + 135x² + 540x³ + 1215 $x^{4}$ + 1458 $x^{5}$ + 729 $x^{6}$

--------------------------------------------------------------------------------------

$(1-3x)^{6}$

= 1. $1^{6}$ $(-3x)^{0}$ + 6. $1^{5}$ $(-3x)^{1}$ + 15. $1^{4}$ $(-3x)^{2}$ + 20. $1^{3}$ $(-3x)^{3}$ + 15.1² $(-3x)^{4}$ + 6. $1^{1}$ $(-3x)^{5}$ + 1. $1^{0}$ $(-3x)^{6}$

= 1 - 18x + 135x² - 540x³ + 1215 $x^{4}$ - 1458 $x^{5}$ + 729 $x^{6}$

----------------------------------------------------------------------------------

Collecting like terms from both expressions

$(1+3x)^{6}$ + $(1-3x)^{6}$

= 2 + 270x² + 2430 $x^{4}$ + 1458 $x^{6}$

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1 5 10 10 5 1

Decreasing powers of 1 from $1^{5}$ to $1^{0}$

Increasing powers of 2x from $(2x)^{0}$ to $(2x)^{5}$

$(1+2x)^{5}$

= 1. $1^{5}$ $(2x)^{0}$ + 5. $1^{4}$ $(2x)^{1}$ + 10. $1^{3}$ $(2x)^{2}$ + 10. $1^{2}$ $(2x)^{3}$ + 5. $1^{1}$ $(2x)^{4}$ + 1. $1^{0}$ $(2x)^{5}$

= 1 + 10x + 40x² + 80x³ + 80 $x^{4}$ + 32 $x^{5}$

8 0

3 years ago

The population of a community is known to increase at a rate proportional to the number of people present at time t. If the popu

Lady bird [3.3K]

Answer:

$P(t) = P(0)e^{0.0693t}$

Step-by-step explanation:

The population of a community is known to increase at a rate proportional to the number of people present at time t.

This means that the population growth is modeled by the following differential equation:

$\frac{dP(t)}{dt} = rP(t)$

Which has the following solution:

$P(t) = P(0)e^{rt}$

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate.

The population has doubled in 10 years

This means that $P(10) = 2P(0)$ . We use this to find r.

$P(t) = P(0)e^{rt}$

$2P(0) = P(0)e^{10r}$

$e^{10r} = 2$

$\ln{e^{10r}} = \ln{2}$

$10r = \ln{2}$

$r = \frac{\ln{2}}{10}$

$r = 0.0693$

So the equation that will estimate the population of the community in t years is:

$P(t) = P(0)e^{0.0693t}$

6 0

3 years ago