Pls help me with my homework I need help

My dad has 4 oak and maple boards from the lumber yard. Each oak board is 3 centimeters thick. Each maple board is 4 centimeters

goldfiish [28.3K]

12 oak boards
60 maple boards

hope this is helpful

4 0

4 years ago

Choose the most appropriate strategy and then use it to solve the quadratic equation.

dsp73

(x-8) ^ 2 = 121
(x-8) = + / - root (121)
x = 8 +/- root (121)
The solutions are:
x1 = 8 + root (121)
x2 = 8 - root (121)
2a ^ 2 = 8a-6
2a ^ 2-8a + 6 = 0
a ^ 2-4a + 3 = 0
(a-1) (a-3) = 0
The solutions are:
a1 = 1
a2 = 3
x ^ 2 + 12x + 36 = 4
x ^ 2 + 12x + 36-4 = 0
x ^ 2 + 12x + 32 = 0
(x + 4) (x + 8) = 0
The solutions are:
x1 = -8
x2 = -4
x ^ 2-x + 30 = 0
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
x = (1 +/- root (1 - 120))) / 2
x = (1 +/- root (-119))) / 2
x = (1 +/- root (119) * i)) / 2
The solutions are:
x1 = (1 + root (119) * i)) / 2
x2 = (1 - root (119) * i)) / 2

7 0

3 years ago

(1 point) the equation for a parabola has the form y=ax2+bx+cy=ax2+bx+c, where aa, bb, and cc are constants and a≠0a≠0. find an

Kryger [21]

A graphing calculator working the quadratic regression problem for these three points gives the equation as

y = 2x² +7x -5

5 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago

Here is a set of signed numbers order them least to greatest: 7, -3, , -0.8, 0.8, -, -2

Aleksandr [31]

Answer:

-3, -2, -0.8, 0.8, 7

7 0

3 years ago

Read 2 more answers