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2 years ago

Pls helllllllllllllpppppppppp mmmmmmmmeeeeee

Mathematics

2 answers:

Masja [62]2 years ago

7 0

Answer:

Hello! answer: 36

Step-by-step explanation:

6 × 7 = 42 3 × 2 = 6 42 - 6 = 36 therefore the area is 36 hope that helps!

ElenaW [278]2 years ago

7 0

Answer:

40cm^2

perimeter:24cm

Step-by-step explanation:

Area: First of all, you can see that the missing height of the shape is 3cm. By adding 4 and 3, you get 7. Then, by multiplying 7 and the length of the shape (6cm), you get 42cm^2. After that, you have to multiply 2cm and 3cm to get the blank area in the bottom right corner to get 6cm^2. Finally, you can subtract 42cm^2(the whole area) by 6cm^2 (the blank area) to get 40cm^2.

Perimeter: Okay! To get the first missing measurement (the one on the left), you have to imagine 4cm being moved to that side. 3cm is parallel to the missing blank space under the 4cm. This means you have to get 7cm (you add 4 and 3). To get the other missing measurement (bottom of the shape) you must see that 6cm would be parallel to the bottom. Since 2cm is also parallel to the bottom, you would have to subtract 2 from 6 to get the bottom of the shape's length, which is 4. Finally, you would have to add 6, 4, 3, 2, 4 and 7 together to get 24cm.

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What is the volume of a circular cylinder with a radius of 6mm and a height of 10mm

Firlakuza [10]

1130.9734
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2 years ago

Need help on this please help

MrRissso [65]

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2 years ago

Giving brainliest and 50 points

kaheart [24]

Answer:

mean: 26

minimum: 0

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median: 27

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max: 45

(all your answers for Aaron are correct as well:)

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1 year ago

Write 9/10 - 3/10 as a mixed number

jeka57 [31]

You do 9/10. Which here it only fits once. So this is 1 1/10. The other one fits 3 times. So its 3 1/10.

If you want the subtraction also done its 1 4/10.

3 0

2 years ago

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin

Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let $\mu$ = true average writing lifetime under controlled conditions

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

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

where, $\bar X$ = sample mean

s = sample standard deviation

n = sample size of pens = 18

n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then we will reject our null hypothesis as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0

2 years ago