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Irina-Kira [14]
2 years ago
14

Help help help help math math

Mathematics
2 answers:
FrozenT [24]2 years ago
8 0
5


??? • 7 = 7 • 5

this means you need an equation that equals 7 • 5 = 35 which means if you switch the numbers you could say 5 • 7
omeli [17]2 years ago
6 0
<h3>Answer:</h3>

5

<h3>Solution:</h3>
  • First, n*7 can be written as 7n.
  • Then, 7*5 is equivalent to 35.
  • So we have
  • 7n=35
  • In order to find the value of n, we need to divide both sides by 7:
  • n=5

Hope it helps.

Do comment if you have any query.

You might be interested in
7.Sabrina wants to deposit $3,066 into savings accounts at three different banks: Bank of the US,
sweet-ann [11.9K]

The amount Sheila deposits in the Bank of US saving account is $511.

Sabrina has a total of $3,066 she wants to deposit. She has three banks that she wants to deposit her money in.

Let, a represent the amount she would deposit she would invest 7/4 bank.

The amount invested in Catch bank = 6 x a = 6a

The amount invested in Bank of US = 20% x ( 6a + a)

= 0.2 x 7a

= 1.4a

The total amount invested in the three banks can be represented with this equation:

1.4a + a + 6a = 3066

In order to determine the amount she would save in the Bank of US, the amount deposited in 7/4 bank has to be determined first.

8.4a = 3066

a = $365

The amount deposited in the Bank of US = 1.4a

= 1.4 x 365

= $511

A similar question was answered here: brainly.com/question/2289204?referrer=searchResults

7 0
3 years ago
What value of x makes the equation 5(4x - 1)=4(2.5x+8) true?
monitta

Answer:

37/10

( I hope this was helpful) >;D

7 0
3 years ago
Read 2 more answers
The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c
Anestetic [448]

Answer:

a) The mean is \mu = 60

b) The standard deviation is \sigma = 9

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when X = 55.5, Z = -0.5

So

Z = \frac{X - \mu}{\sigma}

-0.5 = \frac{55.5 - \mu}{\sigma}

-0.5\sigma = 55.5 - \mu

\mu = 55.5 + 0.5\sigma

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when X = 71.52, Z = 1.28

So

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{71.52 - \mu}{\sigma}

1.28\sigma = 71.52 - \mu

\mu = 71.52 - 1.28\sigma

Since we also have that \mu = 55.5 + 0.5\sigma

55.5 + 0.5\sigma = 71.52 - 1.28\sigma

1.78\sigma = 71.52 - 55.5

\sigma = \frac{(71.52 - 55.5)}{1.78}

\sigma = 9

\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60

Question

The mean is \mu = 60

The standard deviation is \sigma = 9

6 0
3 years ago
A fair die is cast four times. Calculate
svetlana [45]

Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cfrak%7B%20%5Corange%7B%20Question%7D%7D" id="TexFormula1" title=" \large\frak{ \
SVEN [57.7K]

Answer:

\boxed{\sf distance \: between \:  the \: objects \: \: is  \: 2.5 \: cm}

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

A = G \cdot  \frac{m_1.m_2}{ {d}^{2} }

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

<em>Given:</em>

A1 = 2 units

d1= 5 cm

A2 = 8 units

<em>To find:</em>

Distance of separation when the force of attraction is 8 units d2 = ?

<em>Solution:</em>

Substituting the given values in above at each point,

A_1 = G \cdot  \frac{m_1.m_2}{ {d_1}^{2} }  \\ 2 = G \cdot  \frac{m_1.m_2}{ {5}^{2} }  \\  \sf similarly  \: at \:  second \:  point \:  of  \: attraction \\  A_2 = G \cdot  \frac{m_1.m_2}{ {d_2}^{2} }  \\ 8 = G \cdot  \frac{m_1.m_2}{ {d_2}^{2} }  \\  \sf \: \: dividing \:  both  \: of \:  the \:  equation \\  \frac{2}{8}  =  \frac{G \cdot  \frac{m_1.m_2}{ {5}^{2} }}{G \cdot  \frac{m_1.m_2}{ {d_2}^{2} } }  \\  \sf \: G \: m_1 and \: m_2  \: are \:   \: constant  \: hence  \\ \sf  they \:  can  \: be  \: cancelled \:  out  \\  \frac{2}{8}  =  \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } }  \\  \sf rearranging \: above \: equation \\  \frac{1}{4}  =  \frac{{d_2}^{2} }{ {5}^{2} }  \\ {d_2}^{2}  =   \frac{ {5}^{2} }{4 }  \\ {d_2}^{2}  =  \frac{25}{4}   \\ {d_2}^{2}  = 6.25 \\  \sqrt{{d_2}^{2} }  =  \sqrt{6.25}  \\  \boxed{ \sf{d_2} = 2.5 \: cm}

<em>Answer:</em><em> </em><em>the distance between the two </em><em>objects </em><em>is </em><em>2</em><em>.</em><em>5</em><em> </em><em>cm</em><em>, if the attraction between them is 8 </em><em>units.</em>

<em><u>Learn more about universal law of gravitation here brainly.com/question/27244479</u></em>

<em>Thanks </em><em>for </em><em>joining </em><em>brainly </em><em>community</em><em>!</em>

7 0
2 years ago
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