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3 years ago

-9-5(1-7y)=-20show ur work pleasesolve the equations

Mathematics

1 answer:

il63 [147K]3 years ago

8 0

Given equation: -9-5(1-7y)=-20

Step 1) Distributing -5 over (1-7y), we get

-9 -5 +35y = -20.

Step 2) Now, combining like terms (-9-5 =-14) on left side, we get

-14 +35y = -20.

Step 3) Now, we need to get rid -14 form left side.

In order to get rid -14 from left side, we need to add 14 on both sides.

-14+14 +35y = -20+14

35y = -6.

Step 4) Dividing both sides by 35, we get

35y/35 = -6/35.

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3 years ago

At what annual simple interest rate should a loan of Rs.5000 be given to receive the same interest that is received when a loan

IrinaK [193]

Formula :- 

$\sf \: S.I \: = \frac{P.R.T}{100} \\$

Given that :- 

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Solution :- 

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3 years ago

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3 years ago

Read 2 more answers

In an experiment to measure the lifetimes of parts manufactured from a certain aluminum alloy, 73 parts were loaded cyclically u

kari74 [83]

Answer:

0.0094.

Step-by-step explanation:

For the given experiment, let X be the lifetimes of parts manufactured from a certain aluminum alloy.

sample size =73

Sample mean = 784

Sample standard deviation = 120

Let µ represent the mean number of kilocycles to failure for parts of this type.

Null hypothesis, H0 : µ ≤ 750

Alternative hypothesis, H1 : µ > 750.

$z=\dfrac{\overline{X}-\mu}{\frac{s}{\sqrt{n}}}$

where,

$\overline{X}$ is sample mean.

$\mu$ is population mean.

s is sample standard deviation.

n is the sample size.

$z=\dfrac{783-750}{\frac{120}{\sqrt{73}}}$

$z=2.35$

It is a right tailed test because alternative hypothesis is µ > 750. So, P-value is the probability of observing a sample mean greater that 783 or the probability of P(z>2.35).

$P=P(z>2.35)$

$P=1-P(z$

$P=1-0.9906$

$P=0.0094$

Therefore the P-value is 0.0094.

5 0

3 years ago