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3 years ago

I need the answer to number 39.

Mathematics

1 answer:

OLga [1]3 years ago

7 0

To find the number of hours you have to 9 hours and 30 minutes by three because it say how many hours did he play each day and it says three days.
So 9 hours and 30 minutes equal to 570 minutes. Divide 570 three times to get 190. Now how many hours are in 190 minutes? The answer is C. 3 hours and 10 minutes I hope this helps

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How would I do this question using odd co

Liono4ka [1.6K]

Consecutive odd numbres are 2 apart
they numbers are x and x+2
square of first plus 3 times 2nd is 24
first is x
2nd is x+2

x²+3(x+2)=24
easy
x²+3x+6=24
minus 24 both sides
x²+3x-18=0
factor
(x+6)(x-3)=0
set to ezr
x+6=0
x=-6
false, this is even
x-3=0
x=3
x+2=5

the numbers are 3 and 5
9+15=24, yah

3 and 5 are numbers

7 0

3 years ago

Read 2 more answers

Brad bought 70.030 m of chain to make necklaces. He used 0.667 m of the chain to make one necklace.

Margaret [11]

Brad bought 70.030m of this chain to make a necklace, then he used 0.667m of it to make one, and how much he had left, the key word ''used'' and have left'' means subtract so all we have to do is subtract 70.030m minus 0.667m and we get the answer.

Answer: 69.363m of his chain is left.

6 0

3 years ago

A corner store sells two kinds of baked goods: cakes and pies. A cake costs $4 and a pie costs $6. In one day, the store sold 9

Troyanec [42]

Answer:

the number of cakes sell is 5

Step-by-step explanation:

The computation of the number of cakes sold is shown below;

Let us assume

The cake be C

And, the pie cost be P

Given that

There is two types of baked goods sold

The cost of the cake is $4

And, the cost of the pie cost is $64

Now the store sold 9 baked goods for a total of $44

So, the equation would be

C + P = 9

P = 9 - C ...... (1)

4C + 6P = 44............(2)

Now put the value of P in equation 2

4C + 6(9 - C) = 44

4C + 54 - 6C = 44

-2C = -10

C = 5

And, P = 4

hence, the number of cakes sell is 5

8 0

3 years ago

Please help me fast!!

Lynna [10]

Y=x+70
Slope-intercept form

5 0

2 years ago

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

3 years ago