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2 years ago

What is the answer to 1/5 of 55?

Mathematics

1 answer:

dusya [7]2 years ago

6 0

1/5 of 55=55/5= 11. Hope that helps. ((:

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I have $100. I spent $40 on a prize and $45 on food. How do I figure out what % is purchase is from the original amount I had?

Paladinen [302]

You could say that you spent (purchased) 85% of the money that you had.

You spent a total of $85.

If you create the fraction 85/100, that is the same as saying you spent 85% of the money.

8 0

2 years ago

I need the answer to the problem

Nookie1986 [14]

4/9 x 7/4 = 28/36 = 7/9

Answer is 7/9

7 0

2 years ago

Read 2 more answers

Please help me with it this vectors exercise. Calculate:

ohaa [14]

Answer:

ummm

Step-by-step explanation:

5 0

2 years ago

Help: <img src="https://tex.z-dn.net/?f=1.%5C%20%26x%5E4-61x%5E2%2B900%3D0%20%5C%5C%202.%20%5C%20%26%20x%5E4-25x%5E2%2B144%3

yawa3891 [41]

$\bf{1) \ x^4-61x^2+900=0 }$

Applying the factorization method, for example:

$\bf{ 0=(x^2)^2-61(x^2)+900=(x^2-36)(x^2-25)}$

$\bf{\\ &=(x+6)(x-6)(x+5)(x-5); }$

$\bf{ x_1=6 \quad x_2=-6 \quad \ \ \ x_3=5 \quad x_4=5 }$

$\bf{2) \ x^4-25x^2+144=0 }$

Applying the factorization method:

$\bf{0&=(x^2)^2-25(x^2)+144=(x^2-16)(x^2-9) }$

$\bf{ =(x-4)(x+4)(x-3)(x+3); }$

$\bf{ x_1=4 \quad x_2=-4 \quad x_3=3 \quad x_4=-3 }$

7 0

1 year ago

A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar su

dem82 [27]

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

H₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

Hₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a t-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }$

Given:

$\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70$

Thus, the value of the test statistic is 4.70.

7 0

2 years ago