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3 years ago

Sally has saved 41 quarters from washing cars. How many cents does Sally have ?

Mathematics

1 answer:

choli [55]3 years ago

6 0

Answer:

Sally has $10.25 or 1,025 pennies

Step-by-step explanation:

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A luxury liner leaves a port on a bearing of 110 degrees and travels 8.8 miles. It then turns due west and travels 2.4 miles. Ho

myrzilka [38]

Answer:

Distance= 6.6 miles

Bearing= N 62.854°W

Step-by-step explanation:

Let's determine angle b first

Angle b=20° (alternate angles)

Using cosine rule

Let the distance between the liner and the port be x

X² =8.8²+2.4²-2(8.8)(2.4)cos20

X²= 77.44 + 5.76-(39.69)

X²= 43.51

X= √43.51

X= 6.596

X= 6.6 miles

Let's determine the angles within the triangle using sine rule

2.4/sin b = 6.6/sin20

(2.4*sin20)/6.6= sin b

0.1244 = sin b

7.146= b°

Angle c= 180-20-7.146

Angle c= 152.854°

For the bearing

110+7.146= 117.146

180-117.146= 62.854°

Bearing= N 62.854°W

8 0

3 years ago

What is the slope of the line below

Fofino [41]

Answer:

C. $\frac{1}{2}$

Step-by-step explanation:

You can use the formula to find the slope: $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

(-1.5, 1.5) & (1.5, 0)

$\frac{0-(-1.5)}{1.5-(-1.5)} =\\\\\frac{0+1.5}{1.5+1.5} =\\\\\frac{1.5}{3} =\\\\\frac{1}{2}$

The slope is $\frac{1}{2}$

3 0

3 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.

<em>Let </em> $\hat p$ <em> = sample proportion</em>

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

3 years ago

Can y’all explain what I have to do and z help me

zysi [14]

Answer:

Step-by-step explanation:

The answer is D. Both are negative because they're both elevations below 0. However, -7 is higher, or closer to the regular level, because -10 is simply farther down.

5 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago