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2 years ago

What equation could be used to find how many grams of mass is in 15.6 kilograms

Mathematics

1 answer:

Yuki888 [10]2 years ago

7 0

Mass by grams = mass by kilograms *1000
so , mass by grams = 15.6 *1000 = 15600 grams

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Someone please help me with this! Is this a independent or dependent? Explain why.

irga5000 [103]

Answer:

independent

Step-by-step explanation:

Dependent equations graph as the same line. These lines are not the same, so the equations are independent.

___

However, the lines are parallel, so the equations are also inconsistent. There is no solution to this system of equations.

6 0

2 years ago

22. Find the slope and Y intercept of the line 3x+4y-9=0

erik [133]

hi there! here's my answer for you...

Answer:

slope:

m = -3/4

y-intercept:

y = −3x/4 + 9/4

hope this helped. have a good one!

4 0

2 years ago

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the

a_sh-v [17]

Answer:

Probability that the sample proportion will be less than 0.03 is 0.10204.

Step-by-step explanation:

We are given that a courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04.

Also, 469 are sampled.

Let $\hat p$ = sample proportion

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = true proportion = 0.04

n = sample size = 469

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample proportion will be less than 0.03 is given by = P( $\hat p$ < 0.03)

P( $\hat p$ < 0.03) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.03-0.04}{\sqrt{\frac{0.03(1-0.03)}{469} } }$ ) = P(Z < -1.27) = 1 - P(Z $\leq$ 1.27)

= 1 - 0.89796 = 0.10204

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.27 in the z table which has an area of 0.89796.

Therefore, probability that the sample proportion will be less than 0.03 is 0.10204.

5 0

2 years ago

Which is an expression to find the time in hours? d = 38 miles, r = 2 miles/hour A.38*2 B.38/2 C.38+2 D.38-2

Cerrena [4.2K]

Based from background knowledge, I believe that the correct answer among all of the given choices would be B.) 38/2. This would give you 19. If I am mistaken, please correct me. But I am definitely positive that this is the right answer.

I hope this is the answer you were looking for and that it helps!! :)

6 0

2 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

2 years ago