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3 years ago

Solve using a suitable inverse operation:

Mathematics

1 answer:

Olenka [21]3 years ago

7 0

Answer:

a = - 10

Step-by-step explanation:

Given

$\frac{a}{5}$ = - 2

a is divided by 2. The inverse operation to division is multiplication, thus

Multiply both sides by 5 to clear the fraction

a = 5 × - 2 = - 10

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Edna is fishing from a small boat. A fish swimming at the same depth as the hook at the end of her fishing line is 12 feet away

sergeinik [125]

Answer:

16 feet

Step-by-step explanation:

The geometry of the problem can be modeled by a right triangle. The Pythagorean theorem tells us the relationship between the various distances.

EH² +HF² = EF²

EH² + 12² = 20²

EH² = 400 -144

EH = √256 = 16

The hook is 16 feet below Edna.

8 0

2 years ago

Read 2 more answers

Which of the following is equal to the rational expression when x≠-2 or -6? 3(x+2)/(x+6)(x+2)

lara31 [8.8K]

Answer:

3(x+2)=0

3x+6=0

3x=-6

x=-2

for, (x+2)(x+2)=0

x2 +4x+4=0

x=-2 , or x=4

and is

(x+6)(x+2)

4 0

2 years ago

PLEASE HELP making brainliest if correct

nexus9112 [7]

Answer:

im pretty sure its B

Step-by-step explanation:

im so sorry if im wrong tho

6 0

2 years ago

The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.

tatuchka [14]

Answer:

a)Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

b) (5211.05 , 5411.7)

95% lower confidence bound on the mean.

c) The test of hypothesis t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Step-by-step explanation:

Step :-1

Given a random sample of 15 devices is selected in the laboratory.

size of the small sample 'n' = 15

An average life of 5311.4 hours and a sample standard deviation of 220.7 hours.

Average of sample mean (x⁻) = 5311.4 hours

sample standard deviation (S) = 220.7 hours.

Step :- 2

a) Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

Level of significance :- α = 0.95 or 0.05

b) The test statistic

$t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }$

$t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }$

t = 5.826

The degrees of freedom γ= n-1 = 15-1 =14

tabulated value t =1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

calculated value t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Null hypothesis is rejected at 95% confidence on the mean.

C) The 95% of confidence limits

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )$

substitute values and simplification , we get

$(5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )$

(5211.05 , 5411.7)

95% lower confidence bound on the mean.

5 0

3 years ago

Under his cell phone plan, Ian pays a flat cost of $59.50 per month and $5 per gigabyte. He wants to keep his bill under $80 per

inessss [21]

You would start by saying his budget is $80 so you start with 80= he pays an initial fee of $59.50 every month plus $5 per gigabyte so you would set it up like
80=59.50+5x
X being how many gigabytes he uses then you solve 80-59.50 is 20.5 divided by 5 gives you 4.1 but you round down cause you don’t want to pass the limit so therefor the most gigabytes he can use is 4

3 0

3 years ago