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3 years ago

Solve the following linear equations: p - 8 =3p

Mathematics

2 answers:

weqwewe [10]3 years ago

8 0

Answer:

P=4

it's a right answer

AfilCa [17]3 years ago

5 0

Answer:

p = -4

Step-by-step explanation:

p - 8 = 3p

-2p - 8 = 0

-2p = 8

p = -4

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You and your friend are practicing for a rowing competition and want to know how far it is to an island in the Indian River Lago

zepelin [54]

The drawing is missing, so i have attached it.

Answer:

24 yards

Step-by-step explanation:

From the image, we can see that AB and DE are both adjacent sides of their respective triangles.

While, CE and BC are both opposite sides of their respective triangles

Now, from your side, x is the distance to the island.

So, from ratio of similar triangles, we can find x.

Thus;

8/x = 3/9

Cross multiply to get;

8 × 9 = 3x

3x = 72

x = 72/3

x = 24

Thus, distance from your side to the island is 24 yards

6 0

3 years ago

Its worth helping xd

nata0808 [166]

Answer:

$c)\ 2\frac{4}{10}$

Step-by-step explanation:

$5\frac{1}{5}-2\frac{8}{10} \\\\5\frac{2}{10}-2\frac{8}{10} \\\\\frac{52}{10}-\frac{28}{10} \\\\ \frac{24}{10} \\\\2\frac{4}{10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers

Helppppop this is 7th grade math

Dennis_Churaev [7]

B and d are equivalent

3 0

2 years ago

Read 2 more answers

During April it was recorded that there were 19 days when it rained, mostly cloudy on 6 days and sunny on 5 days inference in th

Anika [276]

Answer:

I would defend this

Step-by-step explanation:

This makes sense because April is the rainiest month of the year. It's even shown in the say that April showers bring May flowers.

5 0

2 years ago

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu

m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. $\mu=23.50$

$\sigma=5$

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let $\overline{x}$ be the sample mean.

Formula : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then, the probability that the sample mean is at least $25.00 will be :-

$P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z$

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0

4 years ago