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3 years ago

The math department is putting together an order for new calculators. The

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

4 0

Answer: C More students prefer black Model C3 calculators than white Model A1 calculators.

Step-by-step explanation: I did it

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-1.32+x=1 -1 -1 Not sure about this??

Ipatiy [6.2K]

Answer:

x=2.32

Step-by-step explanation:

-1.32+x=1

add 1.32 to both sides

1.32 cancels itself out

so you are left with x=2.32

7 0

3 years ago

(12f^2-9f+15)-(-4f^2-3f+6) subtract.

german

So what you do is distribute the invisible -1 in front of the second parenthasees set and add like terms
remember
x^2+3x^2=4x^2
-(3+4x-x^2)=-1(3x+4x-x^2)=-3-4x+x^2

so first distribute the negative 1
-1(-4f^2-3f+6)=4f^2+3f-6
add
12f^2-9f+15+4f+3f-6
group like terms
12f^2+4f^2-9f+3f+15-6
add
16f^2-6f+9

aswer is 16f^2-6f+9

8 0

4 years ago

sandra used 2 jars of beads to make 16 necklaces how many jars of beads would she need to make 192 neclaces

garik1379 [7]

Answer:

24 Jars

Step-by-step explanation:

Do 16/2, which gives you eight for every necklace. Then do 192/8 which would give you 24. So, she would need 24 jars for 192 necklaces.

4 0

3 years ago

Read 2 more answers

The function y = f(x) is graphed below. What is the average rate of change of the

Yuliya22 [10]

Don’t know but hope you have a good day

7 0

2 years ago

For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it

amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average mpg = 15.6 mpg

s = sample standard deviation = 1.9 mpg

n = sample of minivans = 100

$\mu$ = population average mpg

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.987 < $t_9_9$ < 1.987) = 0.95 {As the critical value of t at 99 degree

of freedom are -1.987 & 1.987 with P = 2.5%}

P(-1.987 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.6-1.987 \times {\frac{1.9}{\sqrt{100} } }$ , $15.6+1.987 \times {\frac{1.9}{\sqrt{100} } }$ ]

= [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0

3 years ago