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2 years ago

10

Find the solutions to the equation below check all the apply 5x^2+7x-5=0

2 answers:

Sliva [168]2 years ago

6 0

Answer:

e and f

Step-by-step explanation:

there is a great app to use called algebrator it helps me everyday for things like this <3

Sergio [31]2 years ago

6 0

Your 2 solutions would be -5/3 and 1/2. I hope this helps! :)

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The dollar-mart grocery stores sales six bars of soap for $1. how many soaps can a customer buy with $9.

jeka94

Answer:

54

Step-by-step explanation:

6 bars of soap = $1

x bars of soap = $9

We need to multiply both sides of the equation by the same amount so that $9 is on the right.

$1 x 9 = $9

6 x 9 = 54

So, you can buy 54 soaps with $9.

Hope this makes sense. Feel free to ask further if need be.

8 0

1 year ago

Valentin [98]

Answer:

Step-by-step explanation:

the answer is d hope it helps

5 0

2 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

BEST ANSWER WILL GET MARKED BRAINIEST

Sloan [31]

Answer:

1,1

3,3

4,4

5,5

Step-by-step explanation:

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4 0

2 years ago

Read 2 more answers

Simplify the expression -1/2(-5/6 + 1/3)

alexira [117]

Answer: 1/4

Step-by-step explanation:

-

5 0

2 years ago