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2 years ago

6

A fort had enough provisions for 100 men for 8 days. If 60 more men were joined then how many days the provisions will last for?

1 answer:

Grace [21]2 years ago

8 0

Answer:

It would last for 3.2 days.

Step-by-step explanation:

100 men = 8 days

10 men = 0.8 days

60 men = 0.8 x 6 = 4.8 days

Now you subtract the days of 100 men and 60 men

8 - 4.8 = 3.2 days

Hope this helps!

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Solve the equation 24+0.44x=19+1.69x

vichka [17]

Answer: X= 4

Step-by-step explanation:

24+0.44x=19+1.69x exp equation like:

24+44x/100=19+169x/100

Multiply left and right side of equation with 100

2400+44x=1900+169x

2400-1900=169x-44x

500=125x

x=500/125

x=4

7 0

2 years ago

Hello will you help me with this?

Norma-Jean [14]

Answer:

A. 432=18x

B. 22-see explanation

C. 24

D. See if your rounded answer and the actual answer are close.

Step-by-step explanation:

A. He buys a new phone for $432, and pays each month. We will use the variable x, and "x" will represent how many months he pays. It says that he pays $18 per month, so the equation would be 432=18x.

B. We need to isolate x, and to do this we have to divide 432 by 18. To approximate this answer, round 432 to 440 and 18 to 20. Now, divide 440 by 20. We get 22, so this will be the estimation.

C. Divide 432 by 18. We get 24.

E. You can see if it is reasonable if they are close in value.

4 0

3 years ago

If Mark got 6 questions out of 40 incorrect,what percent of the questions did he get correct?

tia_tia [17]

6/40 simplified is 3/20
3/20=15/100
100-15=85
The answer is 85%

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2 years ago

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Evaluate. 58−(14)2= ________

Hunter-Best [27]

Answer:

the answer for 58-(14)2=

its = 30

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3 years ago

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4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a)

Umnica [9.8K]

Answer:

a) The standard deviation of the annual income σₓ = 2045

b)

<em>The calculated value Z = 0.608 < 1.645 at 10 % level of significance</em>

<em>Null hypothesis is accepted </em>

<em>The average annual income is greater than $32,000</em>

c)

<em>The calculated value Z = 1.0977 < 1.96 at 5 % level of significance</em>

<em>Null hypothesis is accepted </em>

<em>The average annual income is equal to $33,000</em>

<em>d) </em>

<em>95% of confidence intervals of the Average annual income</em>

<em>(26 ,746.8 ,34, 763.2)</em>

<u>Step-by-step explanation:</u>

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation = $20,450

a)

The standard deviation of the annual income σₓ = $\frac{S.D}{\sqrt{n} }$

= $\frac{20,450}{\sqrt{100} }= 2045$

b)

Given mean of the Population μ = $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

<u><em>Null Hypothesis:- H₀</em></u>: μ > $32,000

<u><em>Alternative Hypothesis</em></u>:H₁: μ < $32,000

Level of significance α = 0.10

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }$

Z= |-0.608| = 0.608

<em>The calculated value Z = 0.608 < 1.645 at 10 % level of significance</em>

<em>Null hypothesis is accepted </em>

<em>The average annual income is greater than $32,000</em>

c)

Given mean of the Population μ = $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

<u><em>Null Hypothesis:- H₀</em></u>: μ = $33,000

<u><em>Alternative Hypothesis</em></u>:H₁: μ ≠ $33,000

Level of significance α = 0.05

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }$

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

<em>The calculated value Z = 1.0977 < 1.96 at 5 % level of significance</em>

<em>Null hypothesis is accepted </em>

<em>The average annual income is equal to $33,000</em>

<em>d) </em>

<em>95% of confidence intervals is determined by</em>

<em></em> $(x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })$ <em></em>

<em></em> $(30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })$ <em></em>

<em>( 30 755 - 4008.2 , 30 755 +4008.2)</em>

<em>95% of confidence intervals of the Average annual income</em>

<em>(26 ,746.8 ,34, 763.2)</em>

<em></em>

<em></em>

4 0

3 years ago