Determine whether a triangle with these side lengths would be acute, right, or obtuse: 5, 7, and 10;

A cabinet door has a perimeter of 62 inches its area is 228 in.² what are the dimensions of the door

Vlad1618 [11]

P=2(L+W)
A=LW

given
P=62
62=2(L+W)
divide 2
31=L+W
minus W
L=31-W

sub into other one
A=LW
A=(31-W)(W)
228=31W-W^2
times -1
W^2-31W=-228
add 228 both sides
W^2-31W+228=0
factor
what 2 numbers multiply to get 228 and add to get -31
-19 and -12
(W-19)(W-12)=0
set to zero

W-19=0
W=19

W-12=0
W=12

sub back

L=31-W
L=31-12
L=19
or
L=31-19
L=12

the doorway is 12in by 19in

4 0

4 years ago

Solve for d. 1.2+d+0.4=9.7

Advocard [28]

Move everything around. d+1.6=9.7. d=8.1.

3 0

4 years ago

Read 2 more answers

Anyone know? the problem is #16

RoseWind [281]

Because they both have two angles tbat are the same size

3 0

3 years ago

A complete graph of the polynomial function f is provided. Use your knowledge of the domain, range, behavior at each x intercept

Papessa [141]

The graph represents a polynomial graph, and the equation of the polynomial graph is f(x) = -(x + 5)³(x - 2)²

<h3>How to determine the equation?</h3>

From the graph, we have the following observations:

The graph has a turning point at x = 2
The graph changes the factor of its direction at x = -5

This means that the graph has a multiplicity of 2 at x = 2 and a multiplicity of 3 at x = -5

So, we have:

f(x) = a(x + 5)³(x - 2)²

The curve is inverted,

This means that a < 0

Assume a = -1.

Then, we have:

f(x) = -(x + 5)³(x - 2)²

Hence, the equation of the polynomial graph is f(x) = -(x + 5)³(x - 2)²

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

3 0

2 years ago

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

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

$p_v =P(t_{(48)}>1.921)=0.0303$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ 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 mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

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

$df=n_{A}+n_{B}-2=26+24-2=48$

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

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

5 0

3 years ago