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3 years ago

I NEED HELP WITH THEM ALL PLEASE

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

17. 11/20

18. 75cm^2

19. 45

20. a. 93 b. 93 c. 90,94

Step-by-step explanation:

I suggest putting more points for it next time that's 4 questions and saying please makes you sound desperate.

Hope this helped! Have a nice day!

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Makovka662 [10]

10 2/7 ÷ 3/14 First Write The Equation

72/7 ÷ 3/14 Use KCF which is Keep, Change, and Flip

1008/21 Use Division Of Equality

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3 years ago

Rewrite the equation y – 4x = –9 in slope-intercept form. simple explanation and proof

dusya [7]

Answer:

y=4x-9

Step-by-step explanation:

add 4x to both sides to isolate y

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3 years ago

Read 2 more answers

180=112+9x+5. Whats the value of X and why?

vovangra [49]

Answer:x=7

Step-by-step explanation:

180=112+9x+5

Collect like terms

180-112-5=9x

63=9x

Divide both sides by 9

63/9=9x/9

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3 years ago

Please help with this, thank you!

dezoksy [38]

Step-by-step explanation:

$\sqrt{1.8 \times \sin(46 ) \div 7.8 {}^{2} } - 50.8 \\ \sqrt{1.8} \times 0.72 \div 60.84 - 50.8 \\ \sqrt{1.29 \div 10.04 } \\ \sqrt{0.128 } \\ 0.358$

8 0

3 years ago

An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0

2 years ago