Grace is 7 years older than Elena. Grace is also 15 years old.

Jonah’s restaurant bill comes to $25.65 and he leaves a 15% tip. What’s the total costof his restaurant bill?

crimeas [40]

His total bill will be roughly 29.50

6 0

2 years ago

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Orin solved an equation and justified his steps as shown in the table.

sergij07 [2.7K]

Answer:

Step-by-step explanation:

Given the equation as

$\frac{3x}{5} -3=12$

apply multiplication property of equality where you multiply every term by 5

$\frac{3x}{5}*5 -3*5=12*5$

3x-15=60------------------apply addition property of equality

3x-15+15=60+15

3x=75--------------------------appy division property of equality by dividing both sides by 3

3x/3=75/3

x=25

6 0

2 years ago

Determine the value of base x if (211)x = (6A)16

igor_vitrenko [27]

$211_x=2x^2+x+1$
$6A_{16}=106_{10}$

$\implies 2x^2+x+1=106\iff 2x^2+x-105=(x-7)(2x+15)=0$

This has two solutions, $x=7$ and $x=-\dfrac{15}2$ .

4 0

2 years ago

P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball

Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 30, p = 0.626$

So

$\mu = E(X) = np = 30*0.626 = 18.78$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65$

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is $P(X \leq 16 + 0.5) = P(X \leq 16.5)$ , which is the pvalue of Z when X = 16.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.5 - 18.78}{2.65}$

$Z = -0.86$

$Z = -0.86$ has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0

2 years ago

The graph below shows the value of Edna's profits f(t), in dollars, after t months:

umka21 [38]

Since we know that the graph of our quadratic function has x-intercepts at (6,0) and (18,0), $x=6$ and $x=18$ are the zeroes of our quadratic. To find our quadratic we are going to factor each zero backwards and multiply them:
$x=6$
$x-6=0$
$x=18$
$x-18=0$

$(x-6)(x-18)=x^{2}-6x-18x+108$
$=x^{2}-24x+108$

Now that we have our quadratic function, we are going to use the average formula: $m= \frac{f(9)-f(3)}{9-3}$
$where$
$f(9)$ is the function evaluated at the <span>9th month
</span> $f(3)$ is the function evaluated at the 3rd month

$m= \frac{f(9)-f(3)}{9-3}$
$m= \frac{[9^{2}-24(9)+108]-[3^{2}-24(3)+108]}{9-3}$
$m= \frac{-27-45}{6}$
$m= \frac{-72}{6}$
$m=-12$

We can conclude that the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month is: <span>B. −11.63 dollars per month.</span>

6 0

2 years ago

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