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Hoochie [10]
2 years ago
11

Syd chooses two different primes, both of which are greater than $10,$ and multiplies them. The resulting product is less than $

350.$ How many different products could Syd have ended up with?
Mathematics
1 answer:
Mars2501 [29]2 years ago
6 0

Answer:

  10

Step-by-step explanation:

There are 7 primes between 10 and 35: 11, 13, 17, 19, 23, 29, 31.

The product of 11 and any of the others will be less than 350, 6 products.

The product of 13 and any below 26 will be less than 350, 3 more products.

The product of 17 and any below 20 will be less than 350, 1 more product.

There are a total of 10 different products below 350 possible.

_____

11·13, 11·17, 11·19, 11·23, 11·29, 11·31, 13·17, 13·19, 13·23, 17·19

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In quadrilateral QRST, m is 68°, m is (3x + 40)°, and m is (5x − 52)°. What are the measures of , , and ? Write the numerical va
GrogVix [38]

Answer:112,112,68

Step-by-step explanation:

You have the answer in the picture. You only need S and is 180-Q =180-68= 112, so the answer is 112,112,68

6 0
2 years ago
If r(x) = 3x – 1 and s(x) = 2x + 1, which expression is equivalent to (StartFraction r Over s EndFraction) (6)?
Airida [17]

Answer:

\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

Step-by-step explanation:

We know that for any two function f(x) and g(x) ,

\dfrac{f}{g}(x)=\dfrac{f(x)}{g(x)}

Given functions : r(x)=3x-1  and s(x)=2x+1

Then, \dfrac{r}{s}(x)=\dfrac{r(x)}{s(x)}

\Rightarrow\ \dfrac{r}{s}(x)=\dfrac{3x-1}{2x+1}

At x= 6 , we get

\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

The , The expression is equivalent to \dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

When we further simplify it , we get \dfrac{r}{s}(6)=\dfrac{18-1}{12+1}=\dfrac{17}{13}

5 0
2 years ago
What type of proof is used extensively in Geometry?
Nikolay [14]

Answer: two-column

Step-by-step explanation:

7 0
1 year ago
Read 2 more answers
Ginger Smith was seen in Dr. Sampson's office today for an allergic reaction to a prescribed drug. Total charges today are $100.
Anna71 [15]

Answer: Ginger Smith will have to pay $23.00

Step-by-step explanation:

When we talk about a Health Insurance, the Allowed amount A is the maximum amount the Health Insurance company will cover for health care services.  

So, if the health care provider charges more than this allowed amount (as in this situation with Ginger Smith), Ginger will have to pay the difference D, which is calculated by:

D=T-A

T=\$100.00 are the total charges

A=\$77.00 is the allowed amount

D=\$100.00-\$77.00

D=\$23.00 This is the difference Ginger Smith will have to pay

6 0
1 year ago
An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "
FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

\large X_{sentences} the mean of the “sentences” group

\large S_{sentences} the standard deviation of the “sentences” group

\large X_{intentional} the mean of the “intentional” group

\large S_{intentional} the standard deviation of the “intentional” group

Then, we can calculate by using the computer

\large X_{sentences}=28.75  

\large S_{sentences}=3.53553

\large X_{intentional}=31.625

\large S_{intentional}=1.40788

\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}

So, the standard error of the difference is

\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of  95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the \large t_{0.05} which is a value of t such that the area under the Student's t distribution  outside the interval \large [-t_{0.05}, +t_{0.05}] is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

\large t_{0.05}=2.365

Now we can compute our confidence interval

\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0
2 years ago
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