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schepotkina [342]
2 years ago
5

Solve this question: 2x+5is less than or equal to 3x+1,xeN.Want full working please​

Mathematics
1 answer:
IgorLugansk [536]2 years ago
5 0

Answer: 4  ≥ x

Step-by-step explanation:

We want to solve the inequality

2*x + 5 ≥ 3*x + 1

We can solve it in the same way as a normal equation, we need to isolate x in one side of the equation.

I will isolate x in the right side.

2*x + 5 ≥ 3*x + 1

First, we subtract 2*x in both sides:

(2*x + 5) - 2*x  ≥ 3*x + 1 - 2*x

5 ≥ x + 1

Now we subtract 1 in both sides:

5 - 1 ≥ x + 1 - 1

4  ≥ x

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rosijanka [135]

Answer:

57.8

Step-by-step explanation:

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3 years ago
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The advertising and sales model of a company is given by the function S = 2t2 + 100, where t is time in months and t ≥ 0. If adv
kirill [66]

its to shift along the x-axis not the y


7 0
3 years ago
5x-2y=5<br> 2x+3y=21<br><br> How do I find x&amp;y???<br><br><br><br> Please answer asap!!!!
Artist 52 [7]
To solve systems of two equations with two variables (in this case, "x" and "y"), you choose one equation and attempt to solve for one of the variables.

For example, I will try solving for x using the first equation.
5x - 2y = 5
5x = 5 + 2y        [move 2y to other side]
x = (5 + 2y) / 5  [divide both sides by 5]
x = 1 + (.4)y      [simplified form]

Now, I plug this value into the x of the second equation to solve for y.

2x + 3y = 21
2 * (1 + (.4y)) + 3y = 21  [plug in x]
2 + (.8)y + 3y = 21          [distribute 2 to the x value we plugged in]
2 + 3.8y = 21                   [add the y terms together]
3.8y = 19                        [subtract 2]
y = 5                               [divide by 5]

Now that we know the value of y, plug it into either of the original equations to find x. I'll plug it into the first one.
5x - 2y = 5 
5x - 2(5) = 5 [plug in y]
5x - 10 = 5    [2 * 5]
5x = 15         [add 10]
x = 3             [divide by 5]

The final answer is x = 3, y = 5.

4 0
3 years ago
Read 2 more answers
We wish to see if the dial indicating the oven temperature for a certain model of oven is properly calibrated. Four ovens of thi
antoniya [11.8K]

Answer:

a. Standard deviation: 4.082

Standard error: 2.041

b. The 95% confidence interval for the actual temperature is (298.5, 311.5).

Upper bound: 311.5

Lower bound: 298.5

c. Test statistic t=2.45

P-value = 0.092

d. There is no enough evidence to claim that the dial of the oven is not properly calibrated. The actual temperature does not significantly differ from 300 °F.

e. If we use a significance level of 10% (a less rigorous test, in which the null hypothesis is rejected with with less requirements), the conclusion changes and now there is enough evidence to claim that the dial is not properly calibrated.

This happens because now the P-value (0.092) is smaller than the significance level (0.10), given statististical evidence for the claim.

Step-by-step explanation:

The mean and standard deviation of the sample are:

M=\dfrac{1}{4}\sum_{i=1}^{4}(305+310+300+305)\\\\\\ M=\dfrac{1220}{4}=305

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(305-(305))^2+(310-(305))^2+(300-(305))^2+(305-(305))^2]}\\\\\\            s=\sqrt{\dfrac{1}{3}\cdot [(0)+(25)+(25)+(0)]}\\\\\\            s=\sqrt{\dfrac{50}{3}}=\sqrt{16.667}\\\\\\s=4.082

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=305.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM (standard error):

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{4.082}{\sqrt{4}}=\dfrac{4.082}{2}=2.041

The degrees of freedom for this sample size are:

df=n-1=4-1=3

The t-value for a 95% confidence interval and 3 degrees of freedom is t=3.18.

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

MOE=t\cdot s_M=3.18 \cdot 2.041=6.5

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

LL=M-t \cdot s_M = 305-6.5=298.5\\\\UL=M+t \cdot s_M = 305+6.5=311.5

The 95% confidence interval for the actual temperature is (298.5, 311.5).

This is a hypothesis test for the population mean.

The claim is that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

Then, the null and alternative hypothesis are:

H_0: \mu=300\\\\H_a:\mu\neq 300

The significance level is 0.05.

The sample has a size n=4.

The sample mean is M=305.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.028.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.082}{\sqrt{4}}=2.041

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{305-300}{2.041}=\dfrac{5}{2.041}=2.45

The degrees of freedom for this sample size are:

df=n-1=4-1=3

This test is a two-tailed test, with 3 degrees of freedom and t=2.45, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>2.45)=0.092

As the P-value (0.092) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

If the significance level is 10%, the P-value (0.092) is smaller than the significance level (0.1) and the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °C does not significantly differ from 300 °C.

5 0
3 years ago
A class has 32 pupils if 12 are boys what percent are girls
valina [46]
So 12 out of 32=12/32=6/16=3/8

percent means parts out of 100 so
x/100=x%

3/8=0.375/1
0.375/1 times 100/100=37.5/100=37.5%

answer is 37.5%
3 0
3 years ago
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