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2 years ago

Need someone to help me please!

Mathematics

2 answers:

agasfer [191]2 years ago

7 0

Answer:

8 cm² ⇒ 3rd answer is correct

Step-by-step explanation:

Formula to find the area of a triangle is ,

Area = $\frac{1}{2}$ × base × height

They have already given that,

Base = 4 cm

Height = 4 cm

Let us find the area now.

Area = $\frac{1}{2}$ × base × height

Area = $\frac{1}{2}$ × 4 × 4

Area = $\frac{16}{2}$

Area = 8 cm²

Hope this helps you :-)

Let me know if you have any other questions :-)

mylen [45]2 years ago

4 0

Answer:

$\huge\boxed{\sf A = 8\ cm\²}$

Step-by-step explanation:

<u>Given data:</u>

Base = b = 4 cm

Height = h = 4 cm

<u>Required:</u>

Area of triangle = A = ?

<u>Formula:</u>

A = bh / 2

<u>Solution:</u>

A = (4)(4) / 2

A = 16 / 2

A = 8 cm²

$\rule[225]{225}{2}$

Hope this helped!

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Plz help 14 points

ValentinkaMS [17]

1)Rewrite the table:
70, 49, 34.3, 24.01, 11.807 {The original size of the wound =70}

2) write the quotient of each number by the number before & notice the value:

49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^⁽ⁿ⁻¹⁾

3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
or f(n) = y = 70(0.7)ⁿ / 0.7==> f(n) = [(0.7)ⁿ ]/ 100.
This is a decreasing exponential function where the coefficient
raised to n is < 1.
The domain is for all n>= 0.
When n→∞, f(n)→0; For n=0==>f(n) =70. So the range of f(n) is:<=70

7 0

3 years ago

A student on the cross- county team runs 30 minutes a day as a part of her training. Write an equation to describe the relations

KengaRu [80]

Answer:

The answer is below

Step-by-step explanation:

A student on the cross- county team runs 30 minutes a day as a part of her training. Write an equation to describe the relationship between the distance she runs in miles, D, and her running speed, in miles per hour, when she runs at a constant speed of 5.4miles per hour for m minutes, and then at b miles per hour for n minutes

Solution:

Given that the student runs for a total of 30 minutes per day. He runs at 5.4 miles per hour for m minutes, and then at b miles per hour for n minutes.

Hence:

m + n = 30

60 minute = 1 hour. Therefore, m minute = m/60 hour, n minute = n/60 hour.

The distance traveled is the product the speed in miles per hour and the time taken to cover the distance. Distance = speed * time

The total distance ran (D) is:

D = 5.4(m/60) + b(n/60)

6 0

3 years ago

What is the greatest common factor?

dsp73

Answer:

the greatest common factor would be -8w^4x^2

Step-by-step explanation:

lets factor

( $-8w^4x^2(x^2w-1)$

8 0

3 years ago

PLAESE HELP ASAP!!!!

vovikov84 [41]

11 5/8 + 9 1/2= 21 1/8
3/4 - 2/5= 7/20
4 5/6 - 2 1/2= 2 1/3
I think these are right, I apologize if they are wrong.

5 0

2 years ago

Read 2 more answers

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

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

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

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

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

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

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ 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 determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case 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".

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

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

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

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

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

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

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

3 years ago