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Alenkasestr [34]

3 years ago

10

Vivian can grow three plants with every seed packet. With three seed packets how many total plants can Vivian have in her backya

rd?

1 answer:

nasty-shy [4]3 years ago

3 0

Each seed packet can plant 3 plants and she has 3 packets, 3 times 3 is 9

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Find the volume of a cuboid of length 8cm, width 9cm and height 4cm

Varvara68 [4.7K]

Answer:

288 cm³

Step-by-step explanation:

cuboid formular: lengtg×width×height

8 0

1 year ago

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borishaifa [10]

Answer:

1) Appliances are connected in parallel

2) Appliances are connected in parallel

3) Appliances are connected in parallel (because in house , appliances are connected in parallel)

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Disadvantage :- Such circuits can draw more power which leads to high bill charges on electricity. Also such connections need more wiring.

7 0

3 years ago

Pleaseeeee help I’m begging someone I PROMISE. I will mark brainlest

Lerok [7]

Answer:

Step-by-step explanation:

a) x>0 → items can not be negative

b) c(2,000) = 5,000+1.3(2,000) = $7,600

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5 0

3 years ago

The sugar content of the syrup is canned peaches is normally distributed. Assumethe can is designed to have standard deviation 5

andreyandreev [35.5K]

Answer: 0.50477

Step-by-step explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation $\sigma=5$ milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of $s=4.8$ milligrams.

Sample size : n= 10

Test statistic for chi-square : $\chi^2=\dfrac{s^2(n-1)}{\sigma^2}$

i.e. $\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944$

Now, P-value = $P(\chi^2>8.2944)=0.50477$ [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477

8 0

2 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

5 0

2 years ago