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Softa [21]
3 years ago
11

I am really bad at math! So can someone please help me again!

Mathematics
2 answers:
Bumek [7]3 years ago
7 0

Answer:

Fraction: x= 8/5

Decimal: x= 1.6

Mixed Number: x= 1 3/5

IRINA_888 [86]3 years ago
3 0
Ahshsheisidis duduwushduwue whhd
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Look at the triangles shown below.
Sophie [7]

Answer:

Variant c

Step-by-step explanation:

c²=4*9

c=6

You can apply phyphagor

b²=6²+9²=36+81=117

3 \sqrt{13}

4 0
3 years ago
Find the coordinates of the midpoint of line AB given the coordinates: A(-1,<br> 5) B(2,-3)*<br> 
Naily [24]

Answer:

              \bold{M_{AB}\left(\frac12\,,\ 1\right)}

Step-by-step explanation:

The midpoint of AB coordinates are:  M_{AB}\left(x_M\,,\ y_M\right)  where:

x_M=\frac{x_A+x_B}2=\frac{-1+2}2=\frac12\\\\y_M=\frac{y_A+y_B}2=\frac{5+(-3)}2=\frac{2}2=1

6 0
4 years ago
For any parallelogram what is not always true
tresset_1 [31]

Answer:

The diagonals are congruent.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
10(x-5)=9(x+5) Solve for x. Thanks!
VashaNatasha [74]

Check attached image for work; your answer is x = 95.

4 0
4 years ago
Read 2 more answers
Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
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