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alisha [4.7K]
2 years ago
7

What is the area of a rectangle if the length is 12 cm and the width is 7 cm?

Mathematics
2 answers:
Zepler [3.9K]2 years ago
8 0

Answer:

84

Step-by-step explanation:

you would multiply the length versus the width

frozen [14]2 years ago
8 0

Answer:

12×7==========N

Step-by-step explanation:

12×7=84, cause the formula of looking for the area is A=L×W

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Answer:

B

Step-by-step explanation:

it would be shifted up

7 0
4 years ago
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-45 = -9w PLS HELPPPPPPPPPPPP its multipication btw
Dafna1 [17]

Answer: w=5 .

Step-by-step explanation: divide -9 on both sides which gives you 5 .So w=5 .

4 0
3 years ago
Find the slope of the line through (2,-2) and 5,-1)
Kitty [74]

Answer:

m=\frac{1}{3}

Step-by-step explanation:

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(2,\:-2\right),\:\\\left(x_2,\:y_2\right)=\left(5,\:-1\right)\\\\m=\frac{-1-\left(-2\right)}{5-2}\\\\m = \frac{-1+2}{3}\\ \\Simplify\\m = \frac{1}{3}

4 0
4 years ago
An investor puts $2,500 into a life insurance policy that pays 8.5% simple annual interest. If no additional investment is made
KATRIN_1 [288]

Answer:

$4,625

Step-by-step explanation:


7 0
3 years ago
A data set about speed dating includes​ "like" ratings of male dates made by the female dates. The summary statistics are nequal
UNO [17]
<h2>Answer with explanation:</h2>

Given : Sample size n= 189

Sample mean : \overline{x}=7.58

Sample standard deviation : s=1.93

Let \mu be the population mean of "like" ratings of male dates made by the female dates.

As per question ,

Null hypothesis : \mu\geq8.00

Alternative hypothesis : \mu , It means the test is a one-tailed t-test. ( we use t-test when population standard deviation is unknown.)

Test statistic:

t_{stat}=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\=\dfrac{7.58-8.00}{\dfrac{1.93}{\sqrt{189}}}=-2.99

For 0.05 significance and df =188 (df=n-1) p-value = .001582. [By t-table]

Since  .001582< 0.05

Decision: p-value < significance level , that means there is statistical significance, so we reject the null hypothesis.

Conclusion : We support the claim at 5% significance that t the population mean of such ratings is less than 8.00.

4 0
4 years ago
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