Pls pls pls pls pls help me pls thanks

If the longest side is 6.2 cm, and it is twice the length of the shortest, then we can write an equation as 2x=6.2, where x is the short side length.

Then, since a triangle has 3 sides, the medium side length is 14.5-6.2-x.

Hope that helped,

-sirswagger21

4 0

3 years ago

The width and length of a rectangle (in feet)are consecutive odd integers. If the length is increased by 5 feet, the area of the

Tresset [83]

Original:
The width and length of a rectangle are consecutive odd integers
so W = x and L = x + 2

If the length is increased by 5 feet, then new L = x + 2 + 5 = x + 7

A = L x W
60 = (x + 7) x
60 = x^2 + 7x
x^2 + 7x - 60 = 0
(x - 5)(x + 12) = 0
x = 5 and x = -12

From here you have x = W = 5 ft and L = x + 2 = 5 + 2 = 7 feet

Area of original = 5 x 7 = 35

answer
the area of the original rectangle: 35 ft^2

4 0

2 years ago

The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi

Sonja [21]

Answer:

$t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349$

$p_v =P(t_{(72)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=88.8$ represent the sample mean

$s=26.6$ represent the sample standard deviation for the sample

$n=73$ sample size

$\mu_o =93$ represent the value that we want to test

$\alpha=0.05$ 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 true mean i lower than 93 min, the system of hypothesis would be:

Null hypothesis: $\mu \geq 93$

Alternative hypothesis: $\mu < 93$

If we analyze the size for the sample is > 30 but 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:

$t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=73-1=72$

Since is a one side test the p value would be:

$p_v =P(t_{(72)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.

5 0

2 years ago