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2 years ago

Taylor is twice as old as jessi if you add their ages together you get 27 hold old are Taylor and jessi

Mathematics

1 answer:

deff fn [24]2 years ago

6 0

Answer: Taylor is 14 and Jessi is 13

Step-by-step explanation:

Since you can't divide 27 by 2, you would most likely break it in half. SInce the question states that Taylor is still older, she would be 14, and Jessi 13.

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A sphere has a radius of 6 inches. What is the volume of the sphere in terms of π? V = π in.3

Harrizon [31]

Answer: 288pi in3

Step-by-step explanation:

Volume of a sphere= 4/3pir^3

R=6inches

V=4/3pir^3

V=4/3pi(6^3)

V=4/3pi(216)

V=864pi/3

V=288pi in^3

5 0

3 years ago

Read 2 more answers

A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic

Delvig [45]

Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean iron concentration = 0.802 cc/cubic meter

s = sample standard deviation = 0.093

n = number of water samples = 27

$\mu$ = population mean

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.706 < $t_2_6$ < 1.706) = 0.90 {As the critical value of t at 26 degree of

freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.706) = 0.90

P( $-1.706 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.706 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ , $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ]

= [ $0.802 -1.706 \times {\frac{0.093}{\sqrt{27} }$ , $0.802 +1.706 \times {\frac{0.093}{\sqrt{27} }$ ]

= [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

4 0

3 years ago

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

8 0

3 years ago

Read 2 more answers

The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $3

DerKrebs [107]

Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

The parameters in this problem are given as follows:

$\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5$

Item a:

The probability is the p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000, hence:

X = 1250:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1250 - 1140}{77.5}$

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1000 - 1140}{77.5}$

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is one subtracted by the p-value of Z when X = 1250, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the p-value of Z when X = 1220, hence:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1220 - 1140}{77.5}$

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

3 0

2 years ago

The ratio to cats at an animal shelter 3:1. if there are 56 dogs and cats in all, how many more dogs than cats are there??

Black_prince [1.1K]

Answer:

There are 28 more dogs than cats at the animal shelter.

Step-by-step explanation:

If the ratio of dogs to cats at an animal shelter is 3:1, that means there are 3 dogs for every 1 cat.

The way I think about it is I try multiplying every number by 3 and 1, kind of like process of elimination.

After using every number from 1-13, I finally got to 14. Now, 3 x 14 = 42 and 1 x 14 = 14. Add 42 + 14 and you will get 56.

So there are 42 dogs and 14 cats at the animal shelter.

However, it is asking for how many more dogs than cats there are, so you have to subtract 42 - 14 and the answer is 28 more dogs than cats.

6 0

3 years ago