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Yuliya22 [10]
2 years ago
12

Carlos invested part of $3,000 in a 10% certificate account and the rest in a 6% passbook account. The total annual interest fro

m both accounts is $270. How much did he invest at 6%?
Mathematics
1 answer:
xxMikexx [17]2 years ago
6 0
(3000-n)*(10/100)+n*(6/100) = 270
300 - 0.1n + 0.06n = 270
300 - (0.1n - 0.6n) = 270
300 - 0.04n = 270
-0.04n = -30
n = -30/-0.04 = 750
Investment at 6% = 750
Investment at 10% = 3000 - 750 = 2250
And to check just plug it back in so,
(0.06*750)+(0.1*2250) = 270
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A game between two equally skilled players is ended before a winner is declared. Player A needs 3 more points to win, and player
alisha [4.7K]

Answer: player A = 11/16 and player B = 5/16

Step-by-step explanation:

If a coin was to be tossed to determine the winner possible outcomes using arithmetical triangle.

Player A needs 2 points = 11

Player B needs 3 points = 5

Total outcome of tossing a coin = 16

Player A = 11/16 = 0.6875

Player B = 5/16 = 0. 3125

Or

Using the fifth roll of the pascal triangle (2+3) outcome

( 1, 4, 6, 4, 1 )

Addition of the first 3 represent Player A chances of winning = ( 1+ 4 + 6 ) = 11

And the last two = ( 1 + 4 ) = 5 represents the chances of team B winning

Total number of outcome = ( 1 + 4 + 6 + 4 + 1 ) = 16

8 0
1 year ago
Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute
valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust                      Jet red                 Cloudtran

$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

   $=\frac{669.8571}{32}$

  = 20.93304

The F-test  statistics value is :

$F=\frac{s_B^2}{s_W^2}$

  $=\frac{63.55714}{20.93304}$

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

3 0
1 year ago
If GH has endpoints G(-8, 1) and H(4, 5), then the midpoint M of GH lies on the line y=-x+1
Gnoma [55]
The midpoint of GH is (-2, 3), and that point satisfies the equation because

3 = -(-2) + 1
3 = 2 + 1
3 = 3


8 0
2 years ago
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The diagram shows 3 identical circles inside a rectangle.
Leya [2.2K]

one side is 12 as the radius is 2 so x 2 for diameter which is 4. there are 3 circles so times it by 3 which = 12.

the other side is 4 as one circle daimeter is 4.

4 x 12 = 48.

I am not exactly sure how to do the second part but i hope this helps u

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1 year ago
(01.07 MC) Lines BC and ED are parallel. They are intersected by transversal AE, in which point B lies between points and E. The
Brrunno [24]

Answer:

m∠ABC = m∠BED; Corresponding Angles Theorem

Step-by-step explanation:

<u>Given:</u> line BC is parallel to line ED m∠ABC = 70° m∠CED = 30°

<u>Prove:</u> m∠BEC = 40°

Statement Justification

1. line BC is parallel to line ED - Given

2. m∠ABC = 70° - Given

3. m∠CED = 30° - Given

4. m∠BEC + m∠CED = m∠BED - Angle Addition Postulate

5. m∠ABC = m∠BED - Corresponding Angles Theorem

6. m∠BEC + 30° = 70° - Substitution Property of Equality

7. m∠BEC = 40° Subtraction Property of Equality

6 0
1 year ago
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