All categories

3 years ago

20(-1.5r +0.75) answer

Mathematics

2 answers:

vovangra [49]3 years ago

7 0

Answer:

-30r + 15

Step-by-step explanation:

ludmilkaskok [199]3 years ago

5 0

-30r + 15

I hope this is what your looking for

You might be interested in

Select all the correct answers.

Tanya [424]

Answer:

x=24

y=21

Step-by-step explanation:

ok budy mark me as brainlsit

8 0

3 years ago

Read 2 more answers

A professor records the difference between the marks scored by the students in the class test last week and those scored in the

dezoksy [38]

Answer: $(10.82\ , 13.18)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\mu\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 35 , large sample (n>30)

Mean difference : $\overline{x}=12$

Standard deviation : $\sigma=3$

Significance level : $\alpha=- 1-0.999=0.001$

Critical value : $z_{\alpha/2}=z_{0.0005}=3.4807567$

Now, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students will be :-

$12\pm\ (3.4807567)\dfrac{2}{\sqrt{35}}\\\\=12\pm1.18\\\\=(10.82\ , 13.18)$

Hence, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students = $(10.82\ , 13.18)$

4 0

3 years ago

Prove that the diagonals of a parallelogram bisect each other

Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof : $\begin{array}{ |c| c | c | } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In \: \triangle ^{s} \:AOB \: and \: COD } \\ \sf{(i)}& \sf{ \angle \: OAB = \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &\sf{ \angle \: OBA= \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. Proved ✔$

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

$\underbrace{ \overbrace {\mathfrak{Carry \: On \: Learning}}}$ ☂