Pls help i really am confused :/

Area of a triangle is given by the formula A=1/2 bh. The area of the triangle shown to the right is 56 sq. units. Find its base

Westkost [7]

Given

base=2x+6
height=x+4

and
area=56 square units

56=1/2 times (2x+6) times (x+4)
times bot sides by 2
112=(2x+6)(x+4)
expand
112=2x²+14x+24
divide both sides by 2
56=x²+7x+12
minus 56 both sides
0=x²+7x-44
factor
waht 2 numbers mulitply to get -44 and add to get 7
11 and -4
0=(x-4)(x+11)
set to 0

x-4=0
x=4

x+11=0
x=-11
false, measures can't be negative

x=4

height=x+4
height=4+4=8

base=2x+6
base=2(4)+6=8+6=14

x=4 and base=14 and height=8

8 0

3 years ago

What is the answer to 23,450 to the nearest thousands

miss Akunina [59]

Answer:

23,000

Step-by-step explanation:

brainliest

4 0

2 years ago

Wegnerkolmp or someone who can explain maths please I need a proper explanation

umka2103 [35]

Answer: 66 degrees

Explanation:

Check out the attached image below. Figure 1 is the original image without any additions or alterations. Then in figure 2, I extend segment BC to form a line going infinitely in both directions. This line crosses segment DE at point F as shown in the second figure.

Note how angles ABC and DFC are alternate interior angles. Because AB is parallel to DE (given by the arrow markers) this means angle DFC is also 24 degrees

Focus on triangle DFC. This is a right triangle. The 90 degree angle is at C.

So we know that the acute angles x and 24 are complementary. They add to 90. Solve for x

x+24 = 90

x+24-24 = 90-24

x = 66

That is why angle CDE is 66 degrees

6 0

2 years ago

Carmen's golf score is 6 strokes less than Linda's. Their two scores total 156. What is each girls score?

DochEvi [55]

Answer:

Carmen's score is 75 strokes.

Linda's score is 81 strokes.

Step-by-step explanation:

Let the score of Linda be "x".

It is given that Carmen's golf score is 6 strokes less than Linda's.

Thus, Carmen's score is (x-6).

The total score of the girls is

= $x + (x-6) = 2x-6$

The total score is given to be 156.

Thus, the equation becomes,

$2x-6 = 156$

$2x= 162$

$x = 81$

Thus Linda's score is 81 and Carmen's score is (156-81) 75.

5 0

3 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

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

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

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

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

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

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

2 years ago