3,050 divided by 75 anwser

Please help me out if appreciate it :)

12345 [234]

Answer:

51

Step-by-step explanation:

they are congruent so they are equal to eachotjer

4 0

2 years ago

Which piece of additional information can be used to prove △CEA ~ △CDB?

8_murik_8 [283]

<h3>Answer:</h3>

∠BDC and ∠AED are right angles

<h3>Step-by-step explanation:</h3>

Because ∠C ≅ ∠C, the additional bit of information above can be used to show AA similarity.

____

None of the other offered choices says anything about <em>both</em> triangles. In order to show similarity, you need information about corresponding parts of the <em>two</em> triangles. Information about one triangle alone is not sufficient.

5 0

3 years ago

If you tossed a coin 10 times, what is the probability of landing on tails every time

finlep [7]

<span>For a fair coin, the probability of tails is 1/2 or 0.5, so the probability of all tails in 10 tosess is (1/2)^10 = 1/1024 = 0.0009765625 </span>

7 0

3 years ago

Read 2 more answers

Solve for x 73 =(9/10x)+19

Yuliya22 [10]

$73=(\frac{9}{10}x)+19$

To solve x:

1. Substract 19 in both sides of the equation:

$\begin{gathered} 73-19=(\frac{9}{10}x)+19-19 \\ \\ 54=\frac{9}{10}x \end{gathered}$

2. Multiply both sides of the ewuation by 10/9:

$\begin{gathered} 54\cdot\frac{10}{9}=\frac{9}{10}x\cdot\frac{10}{9} \\ \\ \frac{540}{9}=x \\ \\ 60=x \end{gathered}$ <h2>Then, the value of x is 60 (x=60)</h2>

3 0

1 year ago

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 17 Allen's hummingb

Anton [14]

a)

E =0.124
Lower limit =3.026
Upper limit = 3.274

b) Conditions: $\sigma$ is understood to have a regular distribution of weights

c) Interpretation: 80 percent of the time, this range will include the actual average weight of Allen's hummingbirds, hence its reliability may be counted on

d) sample size $n \approx 27$ .

<h3>What conditions are necessary for your calculations?</h3>

Generally,
Here n = 17

T= 3.15

$\sigma=0.40$

confidence level =c = 0.80

Here we will usethe z critical value.

z critical value for (1+c) /2 = (1+0.80)/2

z critical value for (1+c) /2= 0.9

Zc = 1.28

Critical value = 1.28

a) Margin of error (E) :

$E = Zc*\frac{\sigma}{\sqrt{n}}\\\\\E = 1.28*\frac{0.40}{\sqrt{17}}$

E =0.124

Margin of error = E =0.124

Lower limit = x-E

L= 3.15 - 0.124

L= 3.026

Lower limit =3.026

Upper limit = x + E

U= 3.15 + 0.124

U= 3.274

Upper limit = 3.274

The following is a confidence interval with a value of 80 percent for the mean weights of Allen's hummingbirds in the area under study: (3.02,3.28)

b) Conditions: $\sigma$ is understood to have a regular distribution of weights

c) Interpretation: 80 percent of the time, this range will include the actual average weight of Allen's hummingbirds, hence its reliability may be counted on.

d) When is already known, the following equation may be used to determine the appropriate number of samples, n:

$n=(\frac{z_c*\sigma }{E})^2$

Margin of error = 0.13

Sample size (n):

$n= (\frac{z_c*\sigma }{E})^2$

$n=(\frac{ 1.28*0.36}{0.09})^2$

n = 26.21

$n \approx 27$