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3 years ago

Simplify the following. (a) 3(-3 + 5x) - 1 (4 - 4x) (b) 3 squareroot 64 x^15 y^3 -2(-15 e^5 t/30 e^-2 t^-3)^0

Mathematics

1 answer:

Kaylis [27]3 years ago

6 0

Answer:

a. 19x-13 b. $2(32x^{15}y^{3}-1).$

Step-by-step explanation:

a. 3(-3+5x)-1(4-4x) = -9+15x-4+4x = 15x+4x-9-4 = 19x-13.

b. $64x^{15}y^{3}-2(-15e^{5}\frac{t}{30}e^{-2}t^{-3})^{0}$

= $64x^{15}y^{3}-2 = 2(32x^{15}y^{3}-1).$

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Solve the proportion below. 7/x=19

alexgriva [62]

Answer:

The value of x is approximately 0.3684.

Step-by-step explanation:

Given:

$\frac{7}x}=19$

We need to solve the proportion.

Solution:

Now we can see that;

In given equation one number is fraction and other number is whole number.

Now To make whole number into fraction we will divide by 1.

hence the whole number will be equal to;

$19=\frac{19}{1}$

Now the new equation will be;

$\frac{7}{x}=\frac{19}{1}$

Now by cross multiplication we get;

$7\times1=19\times x\\\\7=19x$

Now dividing both side by 19 we get;

$\frac{7}{19}=\frac{19x}{19}\\\\x\approx0.3684$

Hence the value of x is approximately 0.3684.

6 0

3 years ago

Use technology or a z-score table to answer the question.

Alik [6]

Answer:

The second choice: Approximately $65.2\%$ of the pretzel bags here will contain between 225 and 245 pretzels.

Step-by-step explanation:

This explanation uses a z-score table where each $z$ entry has two decimal places.

Let $\mu$ represent the mean of a normal distribution of variable $X$ . Let $\sigma$ be the standard deviation of the distribution. The z-score for the observation $x$ would be:

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question,

$\mu = 240$ .
$\sigma = 9.3$ .

Calculate the z-score for $x_1 = 225$ and $x_2 = 245$ . Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.

$\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}$ .

$\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}$ .

The question is asking for the probability $P(225 \le X \le 245)$ (where $X$ is between two values.) In this case, that's the same as $P(-1.61 \le Z \le 0.54)$ .

Keep in mind that the probabilities on many z-table correspond to probability of $P(Z \le z)$ (where $Z$ is no greater than one value.) Therefore, apply the identity $P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$ to rewrite $P(-1.61 \le Z \le 0.54)$ as the difference between two probabilities:

$P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)$ .

Look up the z-table for $P(Z \le 0.54)$ and $P(Z \le -1.61)$ :

$P(Z \le 0.54)\approx 0.70540$ .
$P(Z \le -1.61) \approx 0.05370$ .

$\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}$ .

3 0

3 years ago

ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2). Find the coordinates of A' and R

shutvik [7]

Answer:

$A'= (2,-1)$ and $R'=(4,-2)$ under this translation.

Step-by-step explanation:

A translation in $R^{2}$ is a mapping T from $R^{2}$ to $R^{2}$ defined by $T(x,y) = (x + v_1,y+v_2)$ , where $v=(v_1,v_2)$ is a fixed vector in $R^{2}$ .

From the problem we know that $T(2,4)=(3,2)$ , so we need to find the values $v_1$ and $v_2$ such that $T(2,4) = (2 + v_1,4+v_2)=(3,2)$ , so $3=2 + v_1$ and $4+v_2=2$ , thus $v_1=1$ and $v_2=2$ .