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grandymaker [24]
3 years ago
14

Kevin is a quarterback on the football team.he completed 36 passes during the season.his second-string replacement,Mark complete

d 2/9 as many passes as Kevin.. How many passes did mark complete?
Mathematics
1 answer:
anastassius [24]3 years ago
3 0
36 × \frac{2}{9}

Since 9 is a factor of 36 you can reduce the 36 to 4 and the 9 to one 

Then just multiply what is left

4 <span>× 2 = 8

So, Mark completed 8 passes.</span>
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1) -3/2 × 4x = -6x
2) -3/2 × 3 = -9/2

Answer: -6x−​9/2
3 0
3 years ago
Juan ran the lemonade stand for 3 more days
kykrilka [37]

Answer:

We need more information to be able to answer your question.

8 0
3 years ago
For this item, any answers that are not whole numbers should be entered as decimals, rounded to the hundredth.
IrinaK [193]

we'll proceed along the same lines as the previous one.

the cost of the boots is 90.10 and that includes 6.25% sales tax, so if "x" is the 100%, then 90.10 is really the 106.25%, because is including the extra 6.25%.


\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 90.10&106.25\\ x&100 \end{array}\implies \cfrac{90.10}{x}=\cfrac{106.25}{100} \\\\\\ 9010=106.25x\implies \cfrac{9010}{106.25}=x\implies 88.8=x


so if that's the cost of each boot, without tax, then the markup will be 88.8 - 50.88 = 37.92.

if the 100% is 88.8, how much is 37.92 off of it in percentage?


\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 88.8&100\\ 37.92&x \end{array}\implies \cfrac{88.8}{37.92}=\cfrac{100}{x} \\\\\\ 88.8x=3792\implies x=\cfrac{3792}{88.8}\implies \stackrel{\%}{42.7}\approx x

6 0
4 years ago
The question is in the picture above ^
Burka [1]

Answer:

the area of this fig is

200.53 {m}^{2}

5 0
3 years ago
The mean salary of actuaries is LaTeX: \mu=\$100,000????=$100,000 and the standard deviation is LaTeX: \sigma=\$36,730????=$36,7
krek1111 [17]

Answer:

The lower limit of 95% confidence interval is 99002.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $100,000

Sample mean, \bar{x} = $111,000

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = $36,730

First, we design the null and the alternate hypothesis

H_{0}: \mu = \$100,000\\H_A: \mu \neq \$100,000

We have to find the lower limit of the 95% confidence interval.

95% Confidence interval:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

111000 \pm 1.96(\dfrac{36730}{\sqrt{36}} )\\\\ = 111000 \pm 11998.4667 \\= (99001.5333,122998.4667)\\ \approx (99002,122999)

The lower limit of 95% confidence interval is 99002.

6 0
4 years ago
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