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UkoKoshka [18]
3 years ago
8

Given that x= -3, what is the value of 2ײ - x + 5?​

Mathematics
2 answers:
RSB [31]3 years ago
6 0

Answer:

 26

Step-by-step explanation:

2x^2 -x+5

x=-3

2(-3)^2 -(-3)+5

2(9)-(-3)+5

18-(-3)+5

negative multiplied by negative is positive so you have

18+3+5= 26

Have a nice day :)

Roman55 [17]3 years ago
3 0

Answer:

26

Step-by-step explanation:

2(9)-(-3)+5

2(9)+3+5

18+8

26

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Given that events A and B are independent with P(A)=0.3P(A)=0.3 and P(A\cap B)=0.27P(A∩B)=0.27, determine the value of P(B)P(B),
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Answer:

0.9

Step-by-step explanation:

Since A and B are independent  hence;

P(A∩B) = P(A)P(B)

Given the following

P(A∩B) = 0.27

P(A) = 0.3

Required

P(B)

Substitute into the formula;

P(B) = P(A∩B)/P(A)

P(B) = 0.27/0.3

P(B) = 0.9

Hence the value of P(B) is 0.9

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2 years ago
50 POINTS A cell phone company charges a monthly base fee for a phone plan with a limit of 500 text messages. A $0.10 fee is cha
serg [7]

Answer:

y = n +.1x

Step-by-step explanation:

y is the total. x is how much over 500 texts she is. n is her base for 500 texts and less.

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Maddie converts z = 8(cos(300°) + isin(300°)) into rectangular form z = a + bi and determines that a = 4 and b = 4 StartRoot 3 E
Anastaziya [24]

Answer:

Option C) The value of a is correct, but the value of b is incorrect.

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4 0
3 years ago
A county environmental agency suspects that the fish in a particular polluted lake have elevated mercury levels. To confirm that
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Answer:

a. The 95% confidence interval for the difference between means is (0.071, 0.389).

b. There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

In the case of the confidence interval, we reach this conclusion because the lower bound is greater than 0. This indicates that, with more than 95% confidence, we can tell that the difference in mercury levels is positive.

In the case of the hypothesis test, we conclude that because the P-value indicates there is a little chance we get that samples if there is no significant difference between the mercury levels. This indicates that the values of mercury in the polluted lake are significantly higher than the unpolluted lake.

Step-by-step explanation:

The table with the data is:

Sample 1 Sample 2

0.580    0.382

0.711      0.276

0.571     0.570

0.666    0.366

0.598

The mean and standard deviation for sample 1 are:

M=\dfrac{1}{5}\sum_{i=1}^{5}(0.58+0.711+0.571+0.666+0.598)\\\\\\ M=\dfrac{3.126}{5}=0.63

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.58-(0.63))^2+...+(0.598-(0.63))^2]}\\\\\\            s=\sqrt{\dfrac{1}{4}\cdot [(0.002)+(0.007)+(0.003)+(0.002)+(0.001)]}\\\\\\            s=\sqrt{\dfrac{0.015}{4}}=\sqrt{0.0037}\\\\\\s=0.061

The mean and standard deviation for sample 2 are:

M=\dfrac{1}{4}\sum_{i=1}^{4}(0.382+0.276+0.57+0.366)\\\\\\ M=\dfrac{1.594}{4}=0.4

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(0.382-(0.4))^2+(0.276-(0.4))^2+(0.57-(0.4))^2+(0.366-(0.4))^2]}\\\\\\            s=\sqrt{\dfrac{1}{3}\cdot [(0)+(0.015)+(0.029)+(0.001)]}\\\\\\            s=\sqrt{\dfrac{0.046}{3}}=\sqrt{0.015}\\\\\\s=0.123

<u>Confidence interval</u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

M_d=M_1-M_2=0.63-0.4=0.23

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07

The critical t-value for a 95% confidence interval is t=2.365.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=2.365 \cdot 0.07=0.159

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 0.23-0.159=0.071\\\\UL=M_d+t \cdot s_{M_d} = 0.23+0.159=0.389

The 95% confidence interval for the difference between means is (0.071, 0.389).

<u>Hypothesis test</u>

This is a hypothesis test for the difference between populations means.

The claim is that the fish in this particular polluted lake have signficantly elevated mercury levels.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

The significance level is 0.05.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

M_d=M_1-M_2=0.63-0.4=0.23

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.23-0}{0.07}=\dfrac{0.23}{0.07}=3.42

The degrees of freedom for this test are:

df=n_1+n_2-1=5+4-2=7

This test is a right-tailed test, with 7 degrees of freedom and t=3.42, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>3.42)=0.006

As the P-value (0.006) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

<u> </u>

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

In the case of the confidence interval, we reach this conclusion because the lower bound is greater than 0. This indicates that, with more than 95% confidence, we can tell that the difference in mercury levels is positive.

In the case of the hypothesis test, we conclude that because the P-value indicates there is a little chance we get that samples if there is no significant difference between the mercury levels. This indicates that the values of mercury in the polluted lake are significantly higher than the unpolluted lake.

7 0
3 years ago
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