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Lunna [17]
3 years ago
15

Bryce left a 20% tip on a $45 dinner bill. What was Bryce's total dinner cost?

Mathematics
1 answer:
marin [14]2 years ago
8 0

Answer:

Answer below hopes it helps

Step-by-step explanation:

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PLEASE HELP, I WILL UP POINTS AND VOTE BRAINLIEST! DONT MAKE ME WASTE MY 88 POINTS.
Allushta [10]

Answer:

Ok im  using a standered form calculator:

1. 9/7x + 1/7 = -8

2. y = -6/5x - 3 =  -11

3. y = 1/3x - 2 = -10

Step-by-step explanation:

I will be glad to help more

6 0
3 years ago
HELP PLZZ ASAP !! a. The value of x is 6 , what is the value of y?
erastovalidia [21]

Answer:

a). y = 4.5

b) Scale factor = \frac{3}{4}

Step-by-step explanation:

a). Since, polygon Q is the scaled copy of polygon P

    Both the figures will be similar.

    Therefore, their corresponding sides will be proportional.

    \frac{x}{y} =\frac{4}{3}

    If x = 6,

    \frac{6}{y} =\frac{4}{3}

    y = \frac{3\times 6}{4}

    y = 4.5

b). Since, polygon Q is the scaled copy of polygon P,

    Polygon Q will be the image of preimage polygon P.

    Scale factor = \frac{\text{Length of a side of the image}}{\text{Length of a side of the preimage}}

                         = \frac{3}{4}

6 0
3 years ago
A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random
Alenkasestr [34]

Answer:

a. Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

b. t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

c. p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2>0

Our notation on this case :

n_1 =78 represent the sample size for group 1

n_2 =21 represent the sample size for group 2

\bar X_1 =92.88 represent the sample mean for the group 1

\bar X_2 =86.90 represent the sample mean for the group 2

s_1=15.34 represent the sample standard deviation for group 1

s_2=8.99 represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom. If we replace the values given we have:

t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

Now we can calculate the degrees of freedom given by:

df=78+21-2=97

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

6 0
3 years ago
Determine if the side lengths could form a triangle 29ft, 38ft, 9ft
snow_tiger [21]

Answer:

yes it would be an obtuse triangle i think

Step-by-step explanation:

7 0
3 years ago
Help binomial distributions
Ymorist [56]

We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is


P(\text{k successed on n trials}) = \binom{n}{k}p^k(1-p)^{n-k}


Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.


We can compute their probability and sum them:\binom{5}{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^2 + \binom{5}{4}\left(\frac{3}{7}\right)^4\left(\frac{4}{7}\right)^1 + \binom{5}{5}\left(\frac{3}{7}\right)^5 \approx 0.36788


So, the answer is about 36.79%

3 0
3 years ago
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