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3 years ago

How do i solve this?

Mathematics

1 answer:

Ierofanga [76]3 years ago

7 0

Step 1: Simplify both sides of the equation.

−36=p−30

Step 2: Flip the equation.

p−30=−36

Step 3: Add 30 to both sides.

p−30+30=−36+30

p=−6

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8. Solve the following equation. w - 2 = - 3

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Answer:

c. w = -1

Step-by-step explanation:

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37 percent as a fraction and a decimal

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Evaluate the expression 3(7 + 4)2 − 24 ÷ 6

solong [7]

Answer:

3(7 + 4)2 − 24 ÷ 6 = 62

Step-by-step explanation:

3(7 + 4)2 − 24 ÷ 6 is the given expression.

Now, by the rule of BODMAS, where B = Bracket, O= of, D = divide,

M = multiplication, A = addition and S = subtraction

we try and solve the following expression in the same order.

Solving the bracket first, we get

3(7 + 4)2 − 24 ÷ 6 = 3(11)2 − 24 ÷ 6 = 66 − 24 ÷ 6

Next, we solve divide,

66 − 24 ÷ 6 = 66 - 4

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Hence, 3(7 + 4)2 − 24 ÷ 6 = 62

6 0

3 years ago

Can you help me answer this

Ipatiy [6.2K]

2149 seats. Since 2149 rounds down to 2100 at the nearest hundred, this is the greatest amount of seats possible in the stadium. If there were 2150 seats, it would round to 2200 seats, so 2149 seats is the correct answer.

Hope this helps!

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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

3 years ago