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3 years ago

Jami went to have a manicure that cost $25. She wanted to tip the technician 20% and

Mathematics

1 answer:

OLga [1]3 years ago

5 0

Answer: $35.75

Step-by-step explanation:

1) $25 for the manicure

2) $25*(0.20) = $5 for the tip

3) $5.75 for the tip

$25 + $5 + $5.75 for the total

$35.75 total

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An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From pre

melamori03 [73]

Answer:

We conclude that a compact microwave oven consumes a mean of more than 250 W.

Step-by-step explanation:

We are given that an appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W with a population standard deviation of 15 W.

They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.

Let $\mu$ = mean power consumption for microwave ovens.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 250 W {means that a compact microwave oven consumes a mean of no more than 250 W}

Alternate Hypothesis, $H_A$ : $\mu$ > 250 W {means that a compact microwave oven consumes a mean of more than 250 W}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean power consumption for ovens = 257.3 W

σ = population standard deviation = 15 W

n = sample of microwave ovens = 20

So, the test statistics = $\frac{257.3-250}{\frac{15}{\sqrt{20} } }$

= 2.176

The value of z test statistics is 2.176.

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistic is more than the critical value of t as 2.176 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that a compact microwave oven consumes a mean of more than 250 W.

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3 years ago

There are two numbers. Their GCF is 5. Their LCM is 150. The first number is 25. What is the second number?

Galina-37 [17]

Answer:

Step-by-step explanation:

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3 years ago

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I'd be more than happy to help. Im a sophmore and Love math

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(01.01 MC)Why is 1 + (−5) equal to −4? I NEED HELP ASAPPPP

miv72 [106K]

Answer:

Hey there!

1+(-5)=1-5

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3 years ago

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Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

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3 years ago