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attashe74 [19]
3 years ago
9

Which of the following is the graph of y=- square root x+1

Mathematics
2 answers:
maks197457 [2]3 years ago
4 0

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

y=-\sqrt{x+1}

Find the intercepts

<u>The y-intercept</u> is the value of y when the value of x is equal to zero

so

For x=0 -----> y=-\sqrt{0+1}=-1

The y-intercept is the point (0,-1)

<u>The x-intercept </u>is the value of x when the value of y is equal to zero

so

For y=0 -----> 0=-\sqrt{x+1} ----->x=-1

The x-intercept is the point (-1,0)

<em>Find the domain</em>

we know that

The radicand must be positive

so

x+1\geq0

x\geq-1

All real numbers greater than or equal to -1

The range is all real numbers less than or equal to -1

using a graphing tool

see the attached figure

Alexeev081 [22]3 years ago
3 0

Answer:

C: third graph  picture is attached.

Step-by-step explanation:

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X is normally distributed with mean 250 and standard deviation 40. what value of x does only the top 15% exceed?
lubasha [3.4K]
The top 15% (85th percentile) is the cutoff value x such that

\mathbb P(X\le x)=\mathbb P\left(\dfrac{X-250}{40}\le\dfrac{x-250}{40}\right)=\mathbb P(Z\le z)=0.85

where z is the corresponding cutoff for the standardized normal distribution. We have z\approx1.0364, and so

\dfrac{x-250}{40}=z\implies x\approx291.456
3 0
3 years ago
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VMariaS [17]

a.0+3

b.3

c.3-4

d.-1

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5 0
2 years ago
The table gives estimates of the world population, in millions, from 1750 to 2000. year population year population 1750 790 1900
Tems11 [23]
The poopulation exponential model is given by

P(t)=P_0e^{kt}

Where, P(t) is the population after year t; Po is the initial population, t is the number of years from the starting year; k is the groth constant.

Given that the population in 1750 is 790 and the population in 1800 is 970, we obtain the population exponential equation as follows:

970=790e^{50k} \\  \\ \Rightarrow e^{50k}=1.228 \\  \\ \Rightarrow 50k=\ln{1.228}=0.2053 \\  \\ \Rightarrow k=0.0041

Thus, the exponential equation using the 1750 and the 1800 population values is P(t)=790e^{0.0041t}

The population of 1900 using the 1750 and the 1800 population values is given by

P(t)=790e^{0.041\times150} \\  \\ =790e^{0.6158}=790(1.8511) \\  \\ =1,462

The population of 1950 using the 1750 and the 1800 population values is given by

P(t)=790e^{0.041\times200} \\  \\ =790e^{0.821}=790(2.2729) \\  \\ =1,796

From the table, it can be seen that the actual figure is greater than the exponential model values.
7 0
3 years ago
A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random
Alenkasestr [34]

Answer:

a. Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

b. t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

c. p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2>0

Our notation on this case :

n_1 =78 represent the sample size for group 1

n_2 =21 represent the sample size for group 2

\bar X_1 =92.88 represent the sample mean for the group 1

\bar X_2 =86.90 represent the sample mean for the group 2

s_1=15.34 represent the sample standard deviation for group 1

s_2=8.99 represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom. If we replace the values given we have:

t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

Now we can calculate the degrees of freedom given by:

df=78+21-2=97

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

6 0
3 years ago
Please help me with these 2 question.
maks197457 [2]

Answer:

Step-by-step explanation:

- sqrt(39) and square root 47

are the limits. The - square root of 39 is smaller than - 6. So the integer to use here is - 6

sqrt (47) = 6.686. Here the square root is larger than the closest integer.

The integer to use is 6

- 6 + 6 = 0

4 0
2 years ago
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