All categories

3 years ago

2+24-76-1756(423)

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

I am think that 2+2*4-7*6-17*56(4*23)

2+8-42-17*56*92 Multiply the Numbers
2+8-42-87584 calculate
Answer -8.7616*10^4

You might be interested in

Step 3: Construct an angle and a perpendicular bisector.

Yuliya22 [10]

Answer:

huh

Step-by-step explanation:

5 0

2 years ago

10 POINTS HELP PLEASE!!!

Liono4ka [1.6K]

Help you with what ?

4 0

3 years ago

28.)Find the value of x.320(3x-10°A.) x = 22.67B.) x = 34.67C.) x = 21.6D.) x = 24.25

katen-ka-za [31]

Since the big triangle is an isosceles one, we conclude that the angles of the left triangle are: 32, 90 and 3x-10. Sinche the sum of this angles have to be 180° we have:

$(3x-10)+32+90=180$

Solving for x, we have:

$\begin{gathered} (3x-10)+32+90=180 \\ 3x-10+32+90=180 \\ 3x=180+10-90-32 \\ 3x=68 \\ x=\frac{68}{3} \\ x=22.67 \end{gathered}$

Therefore, x=22.67 and the answer is A.

7 0

1 year ago

Need help with this answer

ioda

Answer:

(2,2) is your answer :)

Step-by-step explanation:

3 0

3 years ago

Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert

adell [148]

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

$Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}$ $= 60.67$

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: $s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}$

Put value in formula of Standard Deviation,

$s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}$ = 40.75

Step-3: Then, we have to find the critical value by chi-square.

$X_{1-\alpha/2}^{2}=3.82$

$X_{1-\alpha/2}^{2}=21.92$

Then, find the confidence interval which is 95%.

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9$

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9$

5 0

3 years ago